Explanation:
Okay, so I need to find the minimum value of this expression:

$$
\frac{(a + b + c)[(a + b)^2 + (a + b + 4c)^2]}{abc}
$$

where $ a $, $ b $, and $ c $ are positive real numbers. Hmm, let me think about how to approach this. It looks like an optimization problem with three variables. Maybe I can use calculus, like taking partial derivatives and setting them to zero? But that might get complicated with three variables. Alternatively, maybe there's a way to apply inequalities like AM-GM or Cauchy-Schwarz? Let me try to simplify the expression first.

First, let me expand the numerator. Let me denote $ S = a + b + c $, but maybe that's not helpful yet. Let me look at the terms inside the brackets:

$$
(a + b)^2 + (a + b + 4c)^2
$$

Let me compute each term:

First term: $ (a + b)^2 = a^2 + 2ab + b^2 $

Second term: $ (a + b + 4c)^2 $. Let me expand that:

Let me denote $ x = a + b $, so the second term is $ (x + 4c)^2 = x^2 + 8xc + 16c^2 $. Therefore, the entire numerator becomes:

$$
(a + b + c)[(a + b)^2 + (a + b + 4c)^2] = (a + b + c)[(a + b)^2 + (a + b + 4c)^2]
$$

Wait, maybe I can write this as:

Let me compute the sum inside the brackets:

$$
(a + b)^2 + (a + b + 4c)^2 = (a + b)^2 + (a + b)^2 + 8c(a + b) + 16c^2 = 2(a + b)^2 + 8c(a + b) + 16c^2
$$

So the entire numerator is:

$$
(a + b + c)[2(a + b)^2 + 8c(a + b) + 16c^2]
$$

Hmm, maybe I can factor something out here? Let me see. Alternatively, maybe I can make substitutions to reduce the number of variables. Since the expression is symmetric in $ a $ and $ b $, perhaps I can assume $ a = b $? Let me check if that's a valid assumption. If the minimum occurs when $ a = b $, then maybe I can set $ a = b $ to simplify the problem. Let me try that.

Let me set $ a = b $. Let me denote $ a = b = x $, and keep $ c = y $. Then the expression becomes:

First, compute $ a + b + c = 2x + y $

Then compute $ (a + b)^2 + (a + b + 4c)^2 = (2x)^2 + (2x + 4y)^2 = 4x^2 + (2x + 4y)^2 $

Let me compute that:

$ (2x + 4y)^2 = 4x^2 + 16xy + 16y^2 $

Therefore, the sum inside the brackets is:

$ 4x^2 + 4x^2 + 16xy + 16y^2 = 8x^2 + 16xy + 16y^2 $

Therefore, the entire numerator is:

$ (2x + y)(8x^2 + 16xy + 16y^2) $

And the denominator is $ abc = x \cdot x \cdot y = x^2 y $

So the entire expression becomes:

$$
\frac{(2x + y)(8x^2 + 16xy + 16y^2)}{x^2 y}
$$

Let me simplify this. Let me factor numerator:

First, note that 8x² + 16xy + 16y² can be factored as 8x² + 16xy + 16y² = 8(x² + 2xy + 2y²). Wait, let me check:

Wait 8x² + 16xy + 16y² = 8x² + 16xy + 16y². Let me factor 8 out? No, 8x² + 16xy + 16y² = 8x² + 16xy + 16y². Alternatively, factor 8 from the first term and 16 from the others? Not sure. Alternatively, maybe factor 8x² + 16xy + 16y² = 8x² + 16xy + 16y². Let me check if it's a perfect square? Let me see:

Suppose it's (ax + by)^2. Then a²x² + 2abxy + b²y². Comparing coefficients:

a² = 8, 2ab = 16, b² = 16. Let me check if possible. If b² = 16, then b = 4. Then 2ab = 16 => 2a*4 = 16 => a = 2. Then a² = 4, but we need a² = 8. Not matching. So not a perfect square. Maybe factor something else?

Alternatively, maybe factor 8x² + 16xy + 16y² = 8(x² + 2xy + 2y²). Let me check:

8(x² + 2xy + 2y²) = 8x² + 16xy + 16y². Yes, that's correct. So the numerator becomes:

(2x + y) * 8(x² + 2xy + 2y²) = 8(2x + y)(x² + 2xy + 2y²)

Therefore, the entire expression is:

$$
\frac{8(2x + y)(x² + 2xy + 2y²)}{x² y}
$$

Hmm, maybe I can simplify this further. Let me write it as:

8 * [ (2x + y)(x² + 2xy + 2y²) ] / (x² y )

Alternatively, maybe I can make substitution variables. Let me set t = y/x. Since x and y are positive, t is positive. Let me set t = y/x, so y = t x. Then substitute into the expression.

Let me do that. Let me substitute y = t x. Then:

First, compute 2x + y = 2x + t x = x(2 + t)

Next, compute x² + 2xy + 2y²:

x² + 2x(t x) + 2(t x)^2 = x² + 2t x² + 2t² x² = x²(1 + 2t + 2t²)

Therefore, the numerator becomes:

8 * [x(2 + t) * x²(1 + 2t + 2t²)] = 8 * x³ (2 + t)(1 + 2t + 2t²)

Denominator is x² y = x² * t x = t x³

Therefore, the entire expression becomes:

[8 * x³ (2 + t)(1 + 2t + 2t²)] / [t x³] = 8 * (2 + t)(1 + 2t + 2t²)/t

So the expression simplifies to:

$$
8 \cdot \frac{(2 + t)(1 + 2t + 2t^2)}{t}
$$

Now, this is a function of t > 0. So now, the problem reduces to minimizing this expression with respect to t > 0. Let me denote this function as f(t):

$$
f(t) = 8 \cdot \frac{(2 + t)(1 + 2t + 2t^2)}{t}
$$

So I need to find the minimum of f(t) for t > 0. Let me compute this function. Let me first expand the numerator:

First compute (2 + t)(1 + 2t + 2t²). Let me multiply them out:

Multiply 2 by each term in the second polynomial:

2 * 1 = 2

2 * 2t = 4t

2 * 2t² = 4t²

Then multiply t by each term:

t * 1 = t

t * 2t = 2t²

t * 2t² = 2t³

So adding all these terms:

2 + 4t + 4t² + t + 2t² + 2t³ = combine like terms:

Constant term: 2

t terms: 4t + t = 5t

t² terms: 4t² + 2t² = 6t²

t³ term: 2t³

Therefore, the numerator is 2 + 5t + 6t² + 2t³. Therefore, f(t) becomes:

$$
8 \cdot \frac{2 + 5t + 6t^2 + 2t^3}{t} = 8 \cdot \left( \frac{2}{t} + 5 + 6t + 2t^2 \right)
$$

So:

$$
f(t) = 8 \left( 2t^{-1} + 5 + 6t + 2t^2 \right)
$$

Now, to find the minimum of this function, I can take derivative with respect to t, set it to zero, and solve for t. Let me compute f'(t):

First, f(t) = 8*(2/t + 5 + 6t + 2t²). Therefore, derivative:

f’(t) = 8*( -2/t² + 0 + 6 + 4t )

Set derivative equal to zero:

-2/t² + 6 + 4t = 0

Multiply both sides by t² to eliminate denominator:

-2 + 6t² + 4t³ = 0

Rearranged:

4t³ + 6t² - 2 = 0

So, 4t³ + 6t² - 2 = 0. Let me write this equation:

4t³ + 6t² - 2 = 0

Hmm, solving this cubic equation. Let me try to find rational roots using Rational Root Theorem. Possible rational roots are ±1, ±2, ±1/2, ±1/4.

Let me test t = 1: 4 + 6 - 2 = 8 ≠ 0

t = -1: -4 + 6 - 2 = 0. Oh! t = -1 is a root. But since t > 0, we can factor out (t + 1):

Let me perform polynomial division or factorization. Let me write the cubic as:

4t³ + 6t² - 2 = (t + 1)(something). Let me do polynomial division.

Divide 4t³ + 6t² - 2 by (t + 1). Let me set up the division:

Divide 4t³ + 6t² + 0t - 2 by t + 1.

First term: 4t³ / t = 4t². Multiply (t + 1) by 4t²: 4t³ + 4t². Subtract from dividend:

(4t³ + 6t²) - (4t³ + 4t²) = 2t². Bring down the next term: 2t² + 0t.

Next term: 2t² / t = 2t. Multiply (t + 1) by 2t: 2t² + 2t. Subtract:

(2t² + 0t) - (2t² + 2t) = -2t. Bring down the -2: -2t - 2.

Next term: -2t / t = -2. Multiply (t + 1) by -2: -2t - 2. Subtract:

(-2t - 2) - (-2t - 2) = 0. Perfect. So the cubic factors as:

(t + 1)(4t² + 2t - 2) = 0

Therefore, the roots are t = -1 and roots of 4t² + 2t - 2 = 0. Since t > 0, we can ignore t = -1. Let me solve 4t² + 2t - 2 = 0.

Using quadratic formula:

t = [-2 ± sqrt( (2)^2 - 4*4*(-2) )]/(2*4) = [-2 ± sqrt(4 + 32)]/8 = [-2 ± sqrt(36)]/8 = [-2 ± 6]/8

So two solutions:

t = (-2 + 6)/8 = 4/8 = 1/2

t = (-2 - 6)/8 = -8/8 = -1

Again, t = -1 is negative, so the only positive root is t = 1/2. Therefore, critical point at t = 1/2.

Now, we need to check if this critical point is a minimum. Let me check the second derivative or test intervals.

Alternatively, since there's only one critical point for t > 0, and as t approaches 0+ and t approaches infinity, f(t) tends to infinity, so this critical point must be the minimum.

Let me confirm by checking the behavior:

As t approaches 0+, 2/t tends to infinity, so f(t) tends to infinity.

As t approaches infinity, 2t² term dominates, so f(t) ~ 8*(2t²) which tends to infinity. Therefore, the critical point at t = 1/2 must be the minimum.

Therefore, the minimum occurs at t = 1/2. Let me compute f(1/2):

First compute f(t) = 8*(2/t + 5 + 6t + 2t²). Let me compute each term:

t = 1/2:

2/t = 2 / (1/2) = 4

5 remains 5

6t = 6*(1/2) = 3

2t² = 2*(1/2)^2 = 2*(1/4) = 1/2

Therefore, adding them up: 4 + 5 + 3 + 1/2 = 12 + 1/2 = 12.5 = 25/2

Multiply by 8: 8*(25/2) = 100

So the minimum value is 100? Wait, but let me check again:

Wait, f(t) = 8*(2/t + 5 + 6t + 2t²). At t = 1/2:

2/t = 4, 5 is 5, 6t = 3, 2t² = 0.5. So total inside the brackets is 4 + 5 + 3 + 0.5 = 12.5. Multiply by 8: 12.5 * 8 = 100. Yes, 100. So the minimum value is 100?

Wait, but hold on. Wait, this was under the assumption that a = b. But is this the actual minimum for the original problem? Because I assumed a = b. Maybe the minimum occurs when a ≠ b? How can I be sure?

Hmm, this is a common issue in optimization problems. Sometimes assuming symmetry gives the correct answer, but sometimes not. Let me check if the original problem is symmetric in a and b. Let me look back at the original expression:

$$
\frac{(a + b + c)[(a + b)^2 + (a + b + 4c)^2]}{abc}
$$

Yes, the expression is symmetric in a and b. Therefore, the minimum should occur when a = b. Therefore, my assumption is valid. Therefore, the minimum value is 100. Wait, but let me check with specific numbers.

Let me take a = b, c such that t = y/x = 1/2. Since t = c/x = 1/2, so c = x/2. Let me set x = 2, so c = 1. Then a = b = 2, c = 1. Let me compute the original expression:

First compute numerator:

(a + b + c) = 2 + 2 + 1 = 5

Then compute [(a + b)^2 + (a + b + 4c)^2] = (4)^2 + (4 + 4*1)^2 = 16 + (8)^2 = 16 + 64 = 80

Therefore, numerator is 5 * 80 = 400

Denominator is abc = 2 * 2 * 1 = 4

Therefore, the value is 400 / 4 = 100. So that checks out. So with a = b = 2, c = 1, the value is 100. So that seems correct.

But let me try another set of values to see if I can get a lower value. Suppose I take a different ratio. Let me try a = 1, b = 1, c = 0.5. Then t = c / a = 0.5 / 1 = 0.5, same as before. So same as above. Let me check:

a + b + c = 1 + 1 + 0.5 = 2.5

(a + b)^2 + (a + b + 4c)^2 = (2)^2 + (2 + 2)^2 = 4 + 16 = 20

Numerator: 2.5 * 20 = 50

Denominator: 1 * 1 * 0.5 = 0.5

Value: 50 / 0.5 = 100. Same result. So that's consistent.

Alternatively, let me try a different approach. Suppose I don't assume a = b. Let me try to use Lagrange multipliers. Let me consider variables a, b, c > 0. Let me denote the function as:

f(a, b, c) = [(a + b + c)( (a + b)^2 + (a + b + 4c)^2 )] / (abc)

We need to find the minimum of f(a, b, c). Let me take natural logarithm to make differentiation easier? Maybe. Let me denote:

Let me set F(a, b, c) = ln f(a, b, c) = ln(a + b + c) + ln[(a + b)^2 + (a + b + 4c)^2] - ln a - ln b - ln c

Then, take partial derivatives with respect to a, b, c, set them to zero. But this might be complicated, but let me try.

First, compute partial derivative of F with respect to a:

dF/da = [1/(a + b + c)] * [ derivative of (a + b + c) ] + [ derivative of ln(...)] - [1/a] - 0 - 0

Wait, more carefully:

Let me denote:

Let me write F(a, b, c) = ln(a + b + c) + ln[ (a + b)^2 + (a + b + 4c)^2 ] - ln a - ln b - ln c

Therefore, partial derivative with respect to a:

dF/da = [1/(a + b + c)] * (1) + [ (2(a + b) + 2(a + b + 4c)(1) ) / ( (a + b)^2 + (a + b + 4c)^2 ) ] - 1/a

Similarly for partial derivatives with respect to b and c.

But this seems very complicated. However, if we assume that the minimum occurs at a = b, then we can set a = b and check if the derivative with respect to a and b are equal, which they would be. But maybe this is too involved. Alternatively, since we already found a critical point with a = b and verified with specific numbers, and the problem is symmetric in a and b, it's likely that this is indeed the minimum.

Alternatively, let me try to use substitution variables. Let me set x = a + b, y = c. Then, since a and b are positive, x > 0, y > 0. Also, since we have symmetry in a and b, maybe we can set a = b, so x = 2a. But this is similar to what I did before.

Alternatively, let me think of variables substitution. Let me set variables such that u = a + b, v = c. Then, the original expression becomes:

[(u + v)(u² + (u + 4v)^2)] / (abc)

But abc = abv. However, since u = a + b, and we have ab. If we fix u, then ab is maximized when a = b, by AM-GM. Wait, but here we have ab in the denominator. So if we want to minimize the entire expression, which has ab in the denominator, then for fixed u, to minimize the expression, we need to maximize ab. Therefore, for fixed u, maximum ab occurs when a = b = u/2. Therefore, the minimum of the original expression occurs when a = b. Therefore, this confirms that assuming a = b is valid. Therefore, the earlier approach is correct, and the minimum is indeed 100.

Therefore, the answer is 100. Wait, but let me check once more with another example. Suppose I take a = 1, b = 1, c = 0.5. Then as before, value is 100. Suppose I take a = 3, b = 3, c = 1.5. Then:

a + b + c = 3 + 3 + 1.5 = 7.5

(a + b)^2 + (a + b + 4c)^2 = (6)^2 + (6 + 6)^2 = 36 + 144 = 180

Numerator: 7.5 * 180 = 1350

Denominator: 3 * 3 * 1.5 = 13.5

Value: 1350 / 13.5 = 100. Same result. So seems consistent.

Alternatively, let me try a different ratio. Suppose I take t = 1/2, but different a and b. Let me take a = 1, b = 1, c = 0.5. As above. If I take a = 1, b = 2, c = ?

Wait, but if I don't set a = b, then how do I choose c? Let me try. Suppose I take a = 1, b = 2, then let me compute the expression and see if I can get lower than 100.

Let me compute:

First, compute a + b + c = 1 + 2 + c = 3 + c

Compute (a + b)^2 + (a + b + 4c)^2 = (3)^2 + (3 + 4c)^2 = 9 + (3 + 4c)^2

So numerator is (3 + c)[9 + (3 + 4c)^2]

Denominator is abc = 1 * 2 * c = 2c

So the expression is [(3 + c)(9 + (3 + 4c)^2)] / (2c)

Let me compute this for some c. Let me try c = 0.5:

Numerator: (3 + 0.5)[9 + (3 + 2)^2] = 3.5 * [9 + 25] = 3.5 * 34 = 119

Denominator: 2 * 0.5 = 1

Value: 119 / 1 = 119 > 100. So worse.

What if I take c = 1?

Numerator: (3 + 1)[9 + (3 + 4)^2] = 4 * [9 + 49] = 4 * 58 = 232

Denominator: 2 * 1 = 2

Value: 232 / 2 = 116 > 100.

What if I take c = 0.25?

Numerator: 3.25 * [9 + (3 + 1)^2] = 3.25 * [9 + 16] = 3.25 * 25 = 81.25

Denominator: 2 * 0.25 = 0.5

Value: 81.25 / 0.5 = 162.5 > 100.

Alternatively, take c = something else. Let me try to find the minimum for this case. Let me set a = 1, b = 2, and find c that minimizes [(3 + c)(9 + (3 + 4c)^2)] / (2c). Let me denote this as g(c). Let me compute derivative? Maybe too tedious, but let me try.

Alternatively, maybe take c such that the ratio is minimized. But since even with c = 0.5, it's 119, which is higher than 100, and with a = b, we get 100, so seems like 100 is indeed lower. Therefore, this suggests that the minimum is indeed 100.

Alternatively, let me try another case where a ≠ b but maybe with different relations. Suppose I take a = 1, b approaching 0. Let me see what happens. Let me take b approaching 0, then a + b + c ≈ a + c, and (a + b)^2 ≈ a², (a + b + 4c)^2 ≈ (a + 4c)^2. Then the numerator ≈ (a + c)(a² + (a + 4c)^2). Denominator ≈ a * b * c. As b approaches 0, denominator approaches 0, so the whole expression tends to infinity. Therefore, not helpful.

Alternatively, take a very large a and b, but then c? Not sure. Probably the expression would go to infinity as well. Therefore, seems like 100 is indeed the minimum.

Alternatively, let me think of using Hölder's inequality or other inequalities. Let me try to think.

The original expression is:

$$
\frac{(a + b + c)[(a + b)^2 + (a + b + 4c)^2]}{abc}
$$

Let me denote x = a + b, y = c. Then, since a and b are positive, x > 0, y > 0. Also, note that ab ≤ (a + b)^2 / 4 = x² / 4 by AM-GM. Therefore, 1/(ab) ≥ 4 / x². Therefore, the original expression can be written as:

[(x + y)(x² + (x + 4y)^2)] / (ab y) ≥ [(x + y)(x² + (x + 4y)^2)] / ( (x² / 4) y ) = 4(x + y)(x² + (x + 4y)^2) / (x² y )

So, if I can find the minimum of 4(x + y)(x² + (x + 4y)^2) / (x² y ), then that would be a lower bound for the original expression. But since we already found that when a = b, the original expression equals 100, and this lower bound would be?

Let me compute this lower bound expression with x = 2a, y = c. Wait, if a = b, then x = 2a, and y = c. Then, with t = y / x = c / (2a) = 1/2 (from previous case). Let me check:

Let me set x = 2a, y = c. Then, t = y / x = c / (2a) = 1/2. Therefore, substituting into the lower bound expression:

4(x + y)(x² + (x + 4y)^2) / (x² y )

Let me compute:

x + y = 2a + c

x² + (x + 4y)^2 = (2a)^2 + (2a + 4c)^2 = 4a² + (2a + 4c)^2

But since c = (1/2)x = (1/2)(2a) = a. Therefore, c = a. Therefore, substituting c = a:

x + y = 2a + a = 3a

x² + (x + 4y)^2 = 4a² + (2a + 4a)^2 = 4a² + (6a)^2 = 4a² + 36a² = 40a²

Therefore, the lower bound expression becomes:

4*(3a)*(40a²) / ( (2a)^2 * a ) = 4*3a*40a² / (4a² * a ) = (480 a³) / (4a³) ) = 120. Wait, but earlier we had 100. Wait, this seems contradictory. Wait, why?

Because the lower bound is 120, but we have an actual value of 100. Which suggests that my lower bound is not tight. Therefore, my approach here is flawed. Because I used ab ≤ x² / 4, which gives 1/(ab) ≥ 4 / x², but equality holds only when a = b. Therefore, the lower bound is not achievable unless a = b. Therefore, when a = b, the lower bound becomes equal to the original expression. Wait, but in my calculation above, when a = b, c = a, then the lower bound expression is 120, but the actual value is 100. Therefore, there must be a miscalculation here.

Wait, let me check again. Let me re-express the lower bound:

Original expression is [(x + y)(x² + (x + 4y)^2)] / (ab y). Then, since ab ≤ x² / 4, then [(x + y)(x² + (x + 4y)^2)] / (ab y) ≥ [(x + y)(x² + (x + 4y)^2)] / ( (x² / 4) y ) = 4(x + y)(x² + (x + 4y)^2) / (x² y )

But when a = b, then ab = x² / 4, so equality holds. Therefore, the lower bound is 4(x + y)(x² + (x + 4y)^2) / (x² y ). But when a = b and c = a, then x = 2a, y = a. Therefore, substituting:

x = 2a, y = a. Then:

Numerator: 4*(x + y)*(x² + (x + 4y)^2) = 4*(3a)*( (4a²) + (2a + 4a)^2 ) = 4*3a*(4a² + 36a²) = 4*3a*40a² = 480a³

Denominator: x² y = (4a²)*a = 4a³

Therefore, the lower bound is 480a³ / 4a³ = 120. But earlier, when I calculated with a = b = 2, c = 1, the original expression was 100. Therefore, there is a contradiction here. Which means my lower bound approach is wrong? How?

Wait, because when I set a = b, then the original expression is [(x + y)(x² + (x + 4y)^2)] / (ab y) = [(x + y)(x² + (x + 4y)^2)] / ( (x² / 4) y ) = 4(x + y)(x² + (x + 4y)^2) / (x² y ). But in reality, when a = b, the original expression is [(x + y)(x² + (x + 4y)^2)] / (ab y) = [(x + y)(x² + (x + 4y)^2)] / ( (x² / 4) y ) = 4(x + y)(x² + (x + 4y)^2) / (x² y ). But when I calculated with a = b = 2, c = 1, which gives x = 4, y = 1. Wait, no! Wait, earlier I set a = b = 2, c = 1. Then x = a + b = 4, y = c = 1. Therefore, substituting into the lower bound expression:

4*(x + y)*(x² + (x + 4y)^2) / (x² y ) = 4*(4 + 1)*(16 + (4 + 4*1)^2) / (16 * 1) = 4*5*(16 + 64)/16 = 4*5*80 / 16 = 4*5*5 = 100. Ah! There was a miscalculation earlier. Previously, I thought x = 2a, y = a, but in reality, when a = b = 2, c = 1, then x = 4, y = 1. Therefore, substituting into the lower bound expression gives 100, which matches the actual value. Therefore, my previous calculation where I thought x = 2a, y = a was wrong. Because if a = b, then x = 2a, but y = c. If in the case where a = b and c = a, then y = a, so x = 2a, y = a. Then, substituting into the lower bound expression:

4*(x + y)*(x² + (x + 4y)^2) / (x² y ) = 4*(2a + a)*( (2a)^2 + (2a + 4a)^2 ) / ( (2a)^2 * a ) = 4*(3a)*(4a² + 36a²) / (4a² * a ) = 4*3a*40a² / 4a³ = (480a³) / 4a³ = 120. Wait, but in the case where a = b = 2, c = 1, which is not c = a, but c = 1, which is different. So if I set c = a, then with a = b = 2, c = 2, then x = 4, y = 2. Then:

Lower bound expression: 4*(4 + 2)*(16 + (4 + 8)^2) / (16 * 2) = 4*6*(16 + 144)/32 = 4*6*160 / 32 = 4*6*5 = 120. Which matches. But in the case where c = 1, which is different from a, then the lower bound expression gives 100. Therefore, the lower bound is not fixed, but depends on the relation between x and y. Therefore, my previous calculation was correct, and there was no contradiction. Therefore, the lower bound approach is valid, but it's not a fixed number, but depends on variables. Therefore, the minimum of the original expression is indeed 100, achieved when a = b and t = 1/2, which corresponds to c = (1/2)x, where x = a + b. Therefore, with a = b, c = (a + b)/2 = a. Wait, no. Wait, t = y / x = c / (a + b) = 1/2. Therefore, c = (a + b)/2. If a = b, then c = (2a)/2 = a. Therefore, c = a. Therefore, in this case, when a = b = c. Wait, but earlier when I took a = b = 2, c = 1, which is not equal to a. Wait, there's confusion here. Let me clarify.

Earlier, when I set t = y / x = c / (a + b) = 1/2. Therefore, c = (a + b)/2. If a = b, then c = (2a)/2 = a. Therefore, c = a. Therefore, in this case, a = b = c. Wait, but earlier when I took a = b = 2, c = 1, which would mean t = c / (a + b) = 1 / 4, not 1/2. Wait, so there was a mistake here. Earlier, when I set t = y / x = c / x, where x = a + b. But in the substitution where I set t = y / x, then with a = b, x = 2a, y = c, so t = c / (2a). Therefore, when I found t = 1/2, that means c / (2a) = 1/2 => c = a. Therefore, in this case, a = b = c. Wait, but earlier when I took a = b = 2, c = 1, that would correspond to t = 1 / (2*2) = 1/4, not 1/2. Therefore, my previous example where I took a = b = 2, c = 1 was not corresponding to t = 1/2. Wait, but earlier when I calculated with a = b = 2, c = 1, I found the value to be 100. But according to the substitution, when t = 1/2, c = a. Therefore, if I take a = b = 2, c = 2, then t = 2 / (2 + 2) = 2 / 4 = 1/2. Let me compute the value in this case:

a = b = 2, c = 2.

Numerator: (2 + 2 + 2)[(2 + 2)^2 + (2 + 2 + 4*2)^2] = 6 * [16 + (8 + 8)^2] = 6 * [16 + 256] = 6 * 272 = 1632

Denominator: abc = 2 * 2 * 2 = 8

Value: 1632 / 8 = 204. Which is higher than 100. Wait, but earlier when I took a = b = 2, c = 1, I had value 100. How is that possible?

Wait, there must be confusion here. Let me re-examine the substitution. Earlier, when I set t = y / x, where x = a + b, y = c. Then, when I found t = 1/2, that corresponds to c = (1/2)x. Therefore, if x = a + b, then c = (a + b)/2. Therefore, if I set a = b, then x = 2a, so c = (2a)/2 = a. Therefore, c = a. Therefore, in this case, a = b = c. But when I took a = b = 2, c = 1, that would mean c = 1, which is not equal to a. Therefore, that case does not correspond to t = 1/2. Therefore, my previous calculation where I took a = b = 2, c = 1 was not under the assumption that t = 1/2. Therefore, there was a mistake in my previous reasoning. Wait, but earlier when I did the substitution with t = y / x, and found that t = 1/2 gives the minimum, but when I took a = b = 2, c = 1, which is t = 1 / (2 + 2) = 1/4, not 1/2. Therefore, how did I get 100 in that case?

Wait, let me recalculate that case. If a = b = 2, c = 1:

Numerator: (a + b + c) = 2 + 2 + 1 = 5

(a + b)^2 + (a + b + 4c)^2 = (4)^2 + (4 + 4*1)^2 = 16 + 8^2 = 16 + 64 = 80

Numerator total: 5 * 80 = 400

Denominator: abc = 2 * 2 * 1 = 4

Value: 400 / 4 = 100. Correct. But according to the substitution, when t = c / (a + b) = 1 / 4, which is different from t = 1/2. Therefore, how come this case gives the same value as the critical point? There must be some error in my substitution.

Wait, earlier when I did substitution with t = y / x, where x = a + b, y = c, then I set t = y / x, and found that the minimum occurs at t = 1/2. But in this case, with t = 1/4, I still get the value 100. Which suggests that my previous calculation of the critical point was wrong? Or there is some miscalculation.

Wait, let me go back. When I did substitution with t = y / x, then I had:

f(t) = 8*(2 + t)(1 + 2t + 2t²)/t. Then I found critical point at t = 1/2. But when I plug t = 1/4 into this function:

f(1/4) = 8*(2 + 1/4)*(1 + 2*(1/4) + 2*(1/4)^2) / (1/4)

Compute step by step:

First, 2 + 1/4 = 9/4

Next, 1 + 2*(1/4) + 2*(1/16) = 1 + 1/2 + 1/8 = (8/8 + 4/8 + 1/8) = 13/8

Therefore, numerator inside: 9/4 * 13/8 = 117/32

Then, f(t) = 8 * (117/32) / (1/4) = 8 * (117/32) * 4 = 8 * (117/8) = 117. Which is not 100. But earlier, when I took a = b = 2, c = 1, which corresponds to t = 1/4, I got 100. Therefore, there is a contradiction here. Which means there was a mistake in my substitution.

Wait, let me check my substitution again. Earlier, I set a = b = x, c = y. Then, I set t = y / x. Then, I derived the expression f(t) = 8*(2 + t)(1 + 2t + 2t²)/t. But when I took a = b = 2, c = 1, then t = y / x = 1 / 2 = 0.5. Wait, no! Wait, if a = b = x = 2, then x = 2, and c = y = 1. Therefore, t = y / x = 1 / 2 = 0.5. Therefore, t = 1/2. Therefore, in this case, t = 1/2, and f(t) = 100. Which matches. Earlier, I thought that t = 1/4, but that was a mistake. Therefore, my previous calculation where I thought t = 1/4 was wrong. Therefore, the case a = b = 2, c = 1 corresponds to t = 1/2, which is the critical point. Therefore, everything is consistent. Therefore, the minimum value is indeed 100.

Therefore, after careful checking, I can confirm that the minimum value is 100. Therefore, the answer is \boxed{100}.

**Final Answer**
\boxed{100}

To find the minimum value of the expression:

$$
\frac{(a + b + c)[(a + b)^2 + (a + b + 4c)^2]}{abc}
$$

where $ a, b, c > 0 $, we begin by analyzing the structure of the expression. It is symmetric in $ a $ and $ b $, which suggests that the minimum may occur when $ a = b $. This symmetry allows us to simplify the problem by setting $ a = b = x $ and $ c = y $, reducing the number of variables.

---

### Step 1: Substitution and Simplification

Let $ a = b = x $, $ c = y $. Then:

- $ a + b + c = 2x + y $
- $ (a + b)^2 = (2x)^2 = 4x^2 $
- $ (a + b + 4c)^2 = (2x + 4y)^2 = 4x^2 + 16xy + 16y^2 $

So the numerator becomes:

$$
(2x + y)\left[4x^2 + (2x + 4y)^2\right] = (2x + y)\left[4x^2 + 4x^2 + 16xy + 16y^2\right] = (2x + y)(8x^2 + 16xy + 16y^2)
$$

The denominator is:

$$
abc = x \cdot x \cdot y = x^2 y
$$

Thus, the expression becomes:

$$
\frac{(2x + y)(8x^2 + 16xy + 16y^2)}{x^2 y}
$$

---

### Step 2: Introduce a New Variable

Let $ t = \frac{y}{x} $, so $ y = tx $. Substituting into the expression:

- $ 2x + y = 2x + tx = x(2 + t) $
- $ 8x^2 + 16xy + 16y^2 = 8x^2 + 16x(tx) + 16(tx)^2 = x^2(8 + 16t + 16t^2) $

So the expression becomes:

$$
\frac{x(2 + t) \cdot x^2(8 + 16t + 16t^2)}{x^2 \cdot tx} = \frac{x^3(2 + t)(8 + 16t + 16t^2)}{x^3 t} = \frac{(2 + t)(8 + 16t + 16t^2)}{t}
$$

Factor the numerator:

$$
8 + 16t + 16t^2 = 8(1 + 2t + 2t^2)
$$

So the expression becomes:

$$
\frac{(2 + t) \cdot 8(1 + 2t + 2t^2)}{t} = 8 \cdot \frac{(2 + t)(1 + 2t + 2t^2)}{t}
$$

---

### Step 3: Minimize the Function

Let:

$$
f(t) = 8 \cdot \frac{(2 + t)(1 + 2t + 2t^2)}{t}
$$

We now minimize $ f(t) $ for $ t > 0 $. Expanding the numerator:

$$
(2 + t)(1 + 2t + 2t^2) = 2(1 + 2t + 2t^2) + t(1 + 2t + 2t^2) = 2 + 4t + 4t^2 + t + 2t^2 + 2t^3 = 2 + 5t + 6t^2 + 2t^3
$$

So:

$$
f(t) = 8 \cdot \frac{2 + 5t + 6t^2 + 2t^3}{t}
$$

Now, take the derivative of $ f(t) $ to find the critical points. Let:

$$
f(t) = 8 \left( \frac{2}{t} + 5 + 6t + 2t^2 \right)
$$

Then:

$$
f'(t) = 8 \left( -\frac{2}{t^2} + 6 + 4t \right)
$$

Set $ f'(t) = 0 $:

$$
-\frac{2}{t^2} + 6 + 4t = 0 \Rightarrow 4t^3 + 6t^2 - 2 = 0
$$

Solving this cubic equation, we find the positive root is:

$$
t = \frac{1}{2}
$$

---

### Step 4: Evaluate the Minimum

Substitute $ t = \frac{1}{2} $ into the expression:

$$
f\left(\frac{1}{2}\right) = 8 \cdot \frac{(2 + \frac{1}{2})(1 + 2 \cdot \frac{1}{2} + 2 \cdot \left(\frac{1}{2}\right)^2)}{\frac{1}{2}}
$$

Compute each term:

- $ 2 + \frac{1}{2} = \frac{5}{2} $
- $ 1 + 1 + \frac{1}{2} = \frac{5}{2} $
- So the numerator is $ \frac{5}{2} \cdot \frac{5}{2} = \frac{25}{4} $
- Denominator is $ \frac{1}{2} $

Thus:

$$
f\left(\frac{1}{2}\right) = 8 \cdot \frac{25/4}{1/2} = 8 \cdot \frac{25}{2} = 100
$$

---

### Step 5: Verify the Minimum

We can verify that this is indeed the minimum by checking the behavior of $ f(t) $ as $ t \to 0^+ $ and $ t \to \infty $, both of which cause $ f(t) \to \infty $. Therefore, the function has a global minimum at $ t = \frac{1}{2} $, and the minimum value is:

$$
\boxed{100}
$$

[Assistant]
Explanation:
Okay, so I need to find the minimum value of this expression:

$$
\frac{(a + b + c)[(a + b)^2 + (a + b + 4c)^2]}{abc}
$$

where $ a $, $ b $, and $ c $ are positive real numbers. Hmm, let me think about how to approach this. It looks like an optimization problem with three variables. Maybe I can use calculus, like taking partial derivatives and setting them to zero? But that might get complicated with three variables. Alternatively, maybe there's a way to apply inequalities like AM-GM or Cauchy-Schwarz? Let me try to simplify the expression first.

First, let me expand the numerator. Let me denote $ S = a + b + c $, but maybe that's not helpful yet. Let me look at the terms inside the brackets:

$$
(a + b)^2 + (a + b + 4c)^2
$$

Let me compute each term:

First term: $ (a + b)^2 = a^2 + 2ab + b^2 $

Second term: $ (a + b + 4c)^2 $. Let me expand that:

Let me denote $ x = a + b $, so the second term is $ (x + 4c)^2 = x^2 + 8xc + 16c^2 $. Therefore, the entire numerator becomes:

$$
(a + b + c)[(a + b)^2 + (a + b + 4c)^2] = (a + b + c)[(a + b)^2 + (a + b + 4c)^2]
$$

Wait, maybe I can write this as:

Let me compute the sum inside the brackets:

$$
(a + b)^2 + (a + b + 4c)^2 = (a + b)^2 + (a + b)^2 + 8c(a + b) + 16c^2 = 2(a + b)^2 + 8c(a + b) + 16c^2
$$

So the entire numerator is:

$$
(a + b + c)[2(a + b)^2 + 8c(a + b) + 16c^2]
$$

Hmm, maybe I can factor something out here? Let me see. Alternatively, maybe I can make substitutions to reduce the number of variables. Since the expression is symmetric in $ a $ and $ b $, perhaps I can assume $ a = b $? Let me check if that's a valid assumption. If the minimum occurs when $ a = b $, then maybe I can set $ a = b $ to simplify the problem. Let me try that.

Let me set $ a = b $. Let me denote $ a = b = x $, and keep $ c = y $. Then the expression becomes:

First, compute $ a + b + c = 2x + y $

Then compute $ (a + b)^2 + (a + b + 4c)^2 = (2x)^2 + (2x + 4y)^2 = 4x^2 + (2x + 4y)^2 $

Let me compute that:

$ (2x + 4y)^2 = 4x^2 + 16xy + 16y^2 $

Therefore, the sum inside the brackets is:

$ 4x^2 + 4x^2 + 16xy + 16y^2 = 8x^2 + 16xy + 16y^2 $

Therefore, the entire numerator is:

$ (2x + y)(8x^2 + 16xy + 16y^2) $

And the denominator is $ abc = x \cdot x \cdot y = x^2 y $

So the entire expression becomes:

$$
\frac{(2x + y)(8x^2 + 16xy + 16y^2)}{x^2 y}
$$

Let me simplify this. Let me factor numerator:

First, note that 8x² + 16xy + 16y² can be factored as 8x² + 16xy + 16y² = 8(x² + 2xy + 2y²). Wait, let me check:

Wait 8x² + 16xy + 16y² = 8x² + 16xy + 16y². Let me factor 8 out? No, 8x² + 16xy + 16y² = 8x² + 16xy + 16y². Alternatively, factor 8 from the first term and 16 from the others? Not sure. Alternatively, maybe factor 8x² + 16xy + 16y² = 8x² + 16xy + 16y². Let me check if it's a perfect square? Let me see:

Suppose it's (ax + by)^2. Then a²x² + 2abxy + b²y². Comparing coefficients:

a² = 8, 2ab = 16, b² = 16. Let me check if possible. If b² = 16, then b = 4. Then 2ab = 16 => 2a*4 = 16 => a = 2. Then a² = 4, but we need a² = 8. Not matching. So not a perfect square. Maybe factor something else?

Alternatively, maybe factor 8x² + 16xy + 16y² = 8(x² + 2xy + 2y²). Let me check:

8(x² + 2xy + 2y²) = 8x² + 16xy + 16y². Yes, that's correct. So the numerator becomes:

(2x + y) * 8(x² + 2xy + 2y²) = 8(2x + y)(x² + 2xy + 2y²)

Therefore, the entire expression is:

$$
\frac{8(2x + y)(x² + 2xy + 2y²)}{x² y}
$$

Hmm, maybe I can simplify this further. Let me write it as:

8 * [ (2x + y)(x² + 2xy + 2y²) ] / (x² y )

Alternatively, maybe I can make substitution variables. Let me set t = y/x. Since x and y are positive, t is positive. Let me set t = y/x, so y = t x. Then substitute into the expression.

Let me do that. Let me substitute y = t x. Then:

First, compute 2x + y = 2x + t x = x(2 + t)

Next, compute x² + 2xy + 2y²:

x² + 2x(t x) + 2(t x)^2 = x² + 2t x² + 2t² x² = x²(1 + 2t + 2t²)

Therefore, the numerator becomes:

8 * [x(2 + t) * x²(1 + 2t + 2t²)] = 8 * x³ (2 + t)(1 + 2t + 2t²)

Denominator is x² y = x² * t x = t x³

Therefore, the entire expression becomes:

[8 * x³ (2 + t)(1 + 2t + 2t²)] / [t x³] = 8 * (2 + t)(1 + 2t + 2t²)/t

So the expression simplifies to:

$$
8 \cdot \frac{(2 + t)(1 + 2t + 2t^2)}{t}
$$

Now, this is a function of t > 0. So now, the problem reduces to minimizing this expression with respect to t > 0. Let me denote this function as f(t):

$$
f(t) = 8 \cdot \frac{(2 + t)(1 + 2t + 2t^2)}{t}
$$

So I need to find the minimum of f(t) for t > 0. Let me compute this function. Let me first expand the numerator:

First compute (2 + t)(1 + 2t + 2t²). Let me multiply them out:

Multiply 2 by each term in the second polynomial:

2 * 1 = 2

2 * 2t = 4t

2 * 2t² = 4t²

Then multiply t by each term:

t * 1 = t

t * 2t = 2t²

t * 2t² = 2t³

So adding all these terms:

2 + 4t + 4t² + t + 2t² + 2t³ = combine like terms:

Constant term: 2

t terms: 4t + t = 5t

t² terms: 4t² + 2t² = 6t²

t³ term: 2t³

Therefore, the numerator is 2 + 5t + 6t² + 2t³. Therefore, f(t) becomes:

$$
8 \cdot \frac{2 + 5t + 6t^2 + 2t^3}{t} = 8 \cdot \left( \frac{2}{t} + 5 + 6t + 2t^2 \right)
$$

So:

$$
f(t) = 8 \left( 2t^{-1} + 5 + 6t + 2t^2 \right)
$$

Now, to find the minimum of this function, I can take derivative with respect to t, set it to zero, and solve for t. Let me compute f'(t):

First, f(t) = 8*(2/t + 5 + 6t + 2t²). Therefore, derivative:

f’(t) = 8*( -2/t² + 0 + 6 + 4t )

Set derivative equal to zero:

-2/t² + 6 + 4t = 0

Multiply both sides by t² to eliminate denominator:

-2 + 6t² + 4t³ = 0

Rearranged:

4t³ + 6t² - 2 = 0

So, 4t³ + 6t² - 2 = 0. Let me write this equation:

4t³ + 6t² - 2 = 0

Hmm, solving this cubic equation. Let me try to find rational roots using Rational Root Theorem. Possible rational roots are ±1, ±2, ±1/2, ±1/4.

Let me test t = 1: 4 + 6 - 2 = 8 ≠ 0

t = -1: -4 + 6 - 2 = 0. Oh! t = -1 is a root. But since t > 0, we can factor out (t + 1):

Let me perform polynomial division or factorization. Let me write the cubic as:

4t³ + 6t² - 2 = (t + 1)(something). Let me do polynomial division.

Divide 4t³ + 6t² - 2 by (t + 1). Let me set up the division:

Divide 4t³ + 6t² + 0t - 2 by t + 1.

First term: 4t³ / t = 4t². Multiply (t + 1) by 4t²: 4t³ + 4t². Subtract from dividend:

(4t³ + 6t²) - (4t³ + 4t²) = 2t². Bring down the next term: 2t² + 0t.

Next term: 2t² / t = 2t. Multiply (t + 1) by 2t: 2t² + 2t. Subtract:

(2t² + 0t) - (2t² + 2t) = -2t. Bring down the -2: -2t - 2.

Next term: -2t / t = -2. Multiply (t + 1) by -2: -2t - 2. Subtract:

(-2t - 2) - (-2t - 2) = 0. Perfect. So the cubic factors as:

(t + 1)(4t² + 2t - 2) = 0

Therefore, the roots are t = -1 and roots of 4t² + 2t - 2 = 0. Since t > 0, we can ignore t = -1. Let me solve 4t² + 2t - 2 = 0.

Using quadratic formula:

t = [-2 ± sqrt( (2)^2 - 4*4*(-2) )]/(2*4) = [-2 ± sqrt(4 + 32)]/8 = [-2 ± sqrt(36)]/8 = [-2 ± 6]/8

So two solutions:

t = (-2 + 6)/8 = 4/8 = 1/2

t = (-2 - 6)/8 = -8/8 = -1

Again, t = -1 is negative, so the only positive root is t = 1/2. Therefore, critical point at t = 1/2.

Now, we need to check if this critical point is a minimum. Let me check the second derivative or test intervals.

Alternatively, since there's only one critical point for t > 0, and as t approaches 0+ and t approaches infinity, f(t) tends to infinity, so this critical point must be the minimum.

Let me confirm by checking the behavior:

As t approaches 0+, 2/t tends to infinity, so f(t) tends to infinity.

As t approaches infinity, 2t² term dominates, so f(t) ~ 8*(2t²) which tends to infinity. Therefore, the critical point at t = 1/2 must be the minimum.

Therefore, the minimum occurs at t = 1/2. Let me compute f(1/2):

First compute f(t) = 8*(2/t + 5 + 6t + 2t²). Let me compute each term:

t = 1/2:

2/t = 2 / (1/2) = 4

5 remains 5

6t = 6*(1/2) = 3

2t² = 2*(1/2)^2 = 2*(1/4) = 1/2

Therefore, adding them up: 4 + 5 + 3 + 1/2 = 12 + 1/2 = 12.5 = 25/2

Multiply by 8: 8*(25/2) = 100

So the minimum value is 100? Wait, but let me check again:

Wait, f(t) = 8*(2/t + 5 + 6t + 2t²). At t = 1/2:

2/t = 4, 5 is 5, 6t = 3, 2t² = 0.5. So total inside the brackets is 4 + 5 + 3 + 0.5 = 12.5. Multiply by 8: 12.5 * 8 = 100. Yes, 100. So the minimum value is 100?

Wait, but hold on. Wait, this was under the assumption that a = b. But is this the actual minimum for the original problem? Because I assumed a = b. Maybe the minimum occurs when a ≠ b? How can I be sure?

Hmm, this is a common issue in optimization problems. Sometimes assuming symmetry gives the correct answer, but sometimes not. Let me check if the original problem is symmetric in a and b. Let me look back at the original expression:

$$
\frac{(a + b + c)[(a + b)^2 + (a + b + 4c)^2]}{abc}
$$

Yes, the expression is symmetric in a and b. Therefore, the minimum should occur when a = b. Therefore, my assumption is valid. Therefore, the minimum value is 100. Wait, but let me check with specific numbers.

Let me take a = b, c such that t = y/x = 1/2. Since t = c/x = 1/2, so c = x/2. Let me set x = 2, so c = 1. Then a = b = 2, c = 1. Let me compute the original expression:

First compute numerator:

(a + b + c) = 2 + 2 + 1 = 5

Then compute [(a + b)^2 + (a + b + 4c)^2] = (4)^2 + (4 + 4*1)^2 = 16 + (8)^2 = 16 + 64 = 80

Therefore, numerator is 5 * 80 = 400

Denominator is abc = 2 * 2 * 1 = 4

Therefore, the value is 400 / 4 = 100. So that checks out. So with a = b = 2, c = 1, the value is 100. So that seems correct.

But let me try another set of values to see if I can get a lower value. Suppose I take a different ratio. Let me try a = 1, b = 1, c = 0.5. Then t = c / a = 0.5 / 1 = 0.5, same as before. So same as above. Let me check:

a + b + c = 1 + 1 + 0.5 = 2.5

(a + b)^2 + (a + b + 4c)^2 = (2)^2 + (2 + 2)^2 = 4 + 16 = 20

Numerator: 2.5 * 20 = 50

Denominator: 1 * 1 * 0.5 = 0.5

Value: 50 / 0.5 = 100. Same result. So that's consistent.

Alternatively, let me try a different approach. Suppose I don't assume a = b. Let me try to use Lagrange multipliers. Let me consider variables a, b, c > 0. Let me denote the function as:

f(a, b, c) = [(a + b + c)( (a + b)^2 + (a + b + 4c)^2 )] / (abc)

We need to find the minimum of f(a, b, c). Let me take natural logarithm to make differentiation easier? Maybe. Let me denote:

Let me set F(a, b, c) = ln f(a, b, c) = ln(a + b + c) + ln[(a + b)^2 + (a + b + 4c)^2] - ln a - ln b - ln c

Then, take partial derivatives with respect to a, b, c, set them to zero. But this might be complicated, but let me try.

First, compute partial derivative of F with respect to a:

dF/da = [1/(a + b + c)] * [ derivative of (a + b + c) ] + [ derivative of ln(...)] - [1/a] - 0 - 0

Wait, more carefully:

Let me denote:

Let me write F(a, b, c) = ln(a + b + c) + ln[ (a + b)^2 + (a + b + 4c)^2 ] - ln a - ln b - ln c

Therefore, partial derivative with respect to a:

dF/da = [1/(a + b + c)] * (1) + [ (2(a + b) + 2(a + b + 4c)(1) ) / ( (a + b)^2 + (a + b + 4c)^2 ) ] - 1/a

Similarly for partial derivatives with respect to b and c.

But this seems very complicated. However, if we assume that the minimum occurs at a = b, then we can set a = b and check if the derivative with respect to a and b are equal, which they would be. But maybe this is too involved. Alternatively, since we already found a critical point with a = b and verified with specific numbers, and the problem is symmetric in a and b, it's likely that this is indeed the minimum.

Alternatively, let me try to use substitution variables. Let me set x = a + b, y = c. Then, since a and b are positive, x > 0, y > 0. Also, since we have symmetry in a and b, maybe we can set a = b, so x = 2a. But this is similar to what I did before.

Alternatively, let me think of variables substitution. Let me set variables such that u = a + b, v = c. Then, the original expression becomes:

[(u + v)(u² + (u + 4v)^2)] / (abc)

But abc = abv. However, since u = a + b, and we have ab. If we fix u, then ab is maximized when a = b, by AM-GM. Wait, but here we have ab in the denominator. So if we want to minimize the entire expression, which has ab in the denominator, then for fixed u, to minimize the expression, we need to maximize ab. Therefore, for fixed u, maximum ab occurs when a = b = u/2. Therefore, the minimum of the original expression occurs when a = b. Therefore, this confirms that assuming a = b is valid. Therefore, the earlier approach is correct, and the minimum is indeed 100.

Therefore, the answer is 100. Wait, but let me check once more with another example. Suppose I take a = 1, b = 1, c = 0.5. Then as before, value is 100. Suppose I take a = 3, b = 3, c = 1.5. Then:

a + b + c = 3 + 3 + 1.5 = 7.5

(a + b)^2 + (a + b + 4c)^2 = (6)^2 + (6 + 6)^2 = 36 + 144 = 180

Numerator: 7.5 * 180 = 1350

Denominator: 3 * 3 * 1.5 = 13.5

Value: 1350 / 13.5 = 100. Same result. So seems consistent.

Alternatively, let me try a different ratio. Suppose I take t = 1/2, but different a and b. Let me take a = 1, b = 1, c = 0.5. As above. If I take a = 1, b = 2, c = ?

Wait, but if I don't set a = b, then how do I choose c? Let me try. Suppose I take a = 1, b = 2, then let me compute the expression and see if I can get lower than 100.

Let me compute:

First, compute a + b + c = 1 + 2 + c = 3 + c

Compute (a + b)^2 + (a + b + 4c)^2 = (3)^2 + (3 + 4c)^2 = 9 + (3 + 4c)^2

So numerator is (3 + c)[9 + (3 + 4c)^2]

Denominator is abc = 1 * 2 * c = 2c

So the expression is [(3 + c)(9 + (3 + 4c)^2)] / (2c)

Let me compute this for some c. Let me try c = 0.5:

Numerator: (3 + 0.5)[9 + (3 + 2)^2] = 3.5 * [9 + 25] = 3.5 * 34 = 119

Denominator: 2 * 0.5 = 1

Value: 119 / 1 = 119 > 100. So worse.

What if I take c = 1?

Numerator: (3 + 1)[9 + (3 + 4)^2] = 4 * [9 + 49] = 4 * 58 = 232

Denominator: 2 * 1 = 2

Value: 232 / 2 = 116 > 100.

What if I take c = 0.25?

Numerator: 3.25 * [9 + (3 + 1)^2] = 3.25 * [9 + 16] = 3.25 * 25 = 81.25

Denominator: 2 * 0.25 = 0.5

Value: 81.25 / 0.5 = 162.5 > 100.

Alternatively, take c = something else. Let me try to find the minimum for this case. Let me set a = 1, b = 2, and find c that minimizes [(3 + c)(9 + (3 + 4c)^2)] / (2c). Let me denote this as g(c). Let me compute derivative? Maybe too tedious, but let me try.

Alternatively, maybe take c such that the ratio is minimized. But since even with c = 0.5, it's 119, which is higher than 100, and with a = b, we get 100, so seems like 100 is indeed lower. Therefore, this suggests that the minimum is indeed 100.

Alternatively, let me try another case where a ≠ b but maybe with different relations. Suppose I take a = 1, b approaching 0. Let me see what happens. Let me take b approaching 0, then a + b + c ≈ a + c, and (a + b)^2 ≈ a², (a + b + 4c)^2 ≈ (a + 4c)^2. Then the numerator ≈ (a + c)(a² + (a + 4c)^2). Denominator ≈ a * b * c. As b approaches 0, denominator approaches 0, so the whole expression tends to infinity. Therefore, not helpful.

Alternatively, take a very large a and b, but then c? Not sure. Probably the expression would go to infinity as well. Therefore, seems like 100 is indeed the minimum.

Alternatively, let me think of using Hölder's inequality or other inequalities. Let me try to think.

The original expression is:

$$
\frac{(a + b + c)[(a + b)^2 + (a + b + 4c)^2]}{abc}
$$

Let me denote x = a + b, y = c. Then, since a and b are positive, x > 0, y > 0. Also, note that ab ≤ (a + b)^2 / 4 = x² / 4 by AM-GM. Therefore, 1/(ab) ≥ 4 / x². Therefore, the original expression can be written as:

[(x + y)(x² + (x + 4y)^2)] / (ab y) ≥ [(x + y)(x² + (x + 4y)^2)] / ( (x² / 4) y ) = 4(x + y)(x² + (x + 4y)^2) / (x² y )

So, if I can find the minimum of 4(x + y)(x² + (x + 4y)^2) / (x² y ), then that would be a lower bound for the original expression. But since we already found that when a = b, the original expression equals 100, and this lower bound would be?

Let me compute this lower bound expression with x = 2a, y = c. Wait, if a = b, then x = 2a, and y = c. Then, with t = y / x = c / (2a) = 1/2 (from previous case). Let me check:

Let me set x = 2a, y = c. Then, t = y / x = c / (2a) = 1/2. Therefore, substituting into the lower bound expression:

4(x + y)(x² + (x + 4y)^2) / (x² y )

Let me compute:

x + y = 2a + c

x² + (x + 4y)^2 = (2a)^2 + (2a + 4c)^2 = 4a² + (2a + 4c)^2

But since c = (1/2)x = (1/2)(2a) = a. Therefore, c = a. Therefore, substituting c = a:

x + y = 2a + a = 3a

x² + (x + 4y)^2 = 4a² + (2a + 4a)^2 = 4a² + (6a)^2 = 4a² + 36a² = 40a²

Therefore, the lower bound expression becomes:

4*(3a)*(40a²) / ( (2a)^2 * a ) = 4*3a*40a² / (4a² * a ) = (480 a³) / (4a³) ) = 120. Wait, but earlier we had 100. Wait, this seems contradictory. Wait, why?

Because the lower bound is 120, but we have an actual value of 100. Which suggests that my lower bound is not tight. Therefore, my approach here is flawed. Because I used ab ≤ x² / 4, which gives 1/(ab) ≥ 4 / x², but equality holds only when a = b. Therefore, the lower bound is not achievable unless a = b. Therefore, when a = b, the lower bound becomes equal to the original expression. Wait, but in my calculation above, when a = b, c = a, then the lower bound expression is 120, but the actual value is 100. Therefore, there must be a miscalculation here.

Wait, let me check again. Let me re-express the lower bound:

Original expression is [(x + y)(x² + (x + 4y)^2)] / (ab y). Then, since ab ≤ x² / 4, then [(x + y)(x² + (x + 4y)^2)] / (ab y) ≥ [(x + y)(x² + (x + 4y)^2)] / ( (x² / 4) y ) = 4(x + y)(x² + (x + 4y)^2) / (x² y )

But when a = b, then ab = x² / 4, so equality holds. Therefore, the lower bound is 4(x + y)(x² + (x + 4y)^2) / (x² y ). But when a = b and c = a, then x = 2a, y = a. Therefore, substituting:

x = 2a, y = a. Then:

Numerator: 4*(x + y)*(x² + (x + 4y)^2) = 4*(3a)*( (4a²) + (2a + 4a)^2 ) = 4*3a*(4a² + 36a²) = 4*3a*40a² = 480a³

Denominator: x² y = (4a²)*a = 4a³

Therefore, the lower bound is 480a³ / 4a³ = 120. But earlier, when I calculated with a = b = 2, c = 1, the original expression was 100. Therefore, there is a contradiction here. Which means my lower bound approach is wrong? How?

Wait, because when I set a = b, then the original expression is [(x + y)(x² + (x + 4y)^2)] / (ab y) = [(x + y)(x² + (x + 4y)^2)] / ( (x² / 4) y ) = 4(x + y)(x² + (x + 4y)^2) / (x² y ). But in reality, when a = b, the original expression is [(x + y)(x² + (x + 4y)^2)] / (ab y) = [(x + y)(x² + (x + 4y)^2)] / ( (x² / 4) y ) = 4(x + y)(x² + (x + 4y)^2) / (x² y ). But when I calculated with a = b = 2, c = 1, which gives x = 4, y = 1. Wait, no! Wait, earlier I set a = b = 2, c = 1. Then x = a + b = 4, y = c = 1. Therefore, substituting into the lower bound expression:

4*(x + y)*(x² + (x + 4y)^2) / (x² y ) = 4*(4 + 1)*(16 + (4 + 4*1)^2) / (16 * 1) = 4*5*(16 + 64)/16 = 4*5*80 / 16 = 4*5*5 = 100. Ah! There was a miscalculation earlier. Previously, I thought x = 2a, y = a, but in reality, when a = b = 2, c = 1, then x = 4, y = 1. Therefore, substituting into the lower bound expression gives 100, which matches the actual value. Therefore, my previous calculation where I thought x = 2a, y = a was wrong. Because if a = b, then x = 2a, but y = c. If in the case where a = b and c = a, then y = a, so x = 2a, y = a. Then, substituting into the lower bound expression:

4*(x + y)*(x² + (x + 4y)^2) / (x² y ) = 4*(2a + a)*( (2a)^2 + (2a + 4a)^2 ) / ( (2a)^2 * a ) = 4*(3a)*(4a² + 36a²) / (4a² * a ) = 4*3a*40a² / 4a³ = (480a³) / 4a³ = 120. Wait, but in the case where a = b = 2, c = 1, which is not c = a, but c = 1, which is different. So if I set c = a, then with a = b = 2, c = 2, then x = 4, y = 2. Then:

Lower bound expression: 4*(4 + 2)*(16 + (4 + 8)^2) / (16 * 2) = 4*6*(16 + 144)/32 = 4*6*160 / 32 = 4*6*5 = 120. Which matches. But in the case where c = 1, which is different from a, then the lower bound expression gives 100. Therefore, the lower bound is not fixed, but depends on the relation between x and y. Therefore, my previous calculation was correct, and there was no contradiction. Therefore, the lower bound approach is valid, but it's not a fixed number, but depends on variables. Therefore, the minimum of the original expression is indeed 100, achieved when a = b and t = 1/2, which corresponds to c = (1/2)x, where x = a + b. Therefore, with a = b, c = (a + b)/2 = a. Wait, no. Wait, t = y / x = c / (a + b) = 1/2. Therefore, c = (a + b)/2. If a = b, then c = (2a)/2 = a. Therefore, c = a. Therefore, in this case, when a = b = c. Wait, but earlier when I took a = b = 2, c = 1, which is not equal to a. Wait, there's confusion here. Let me clarify.

Earlier, when I set t = y / x = c / (a + b) = 1/2. Therefore, c = (a + b)/2. If a = b, then c = (2a)/2 = a. Therefore, c = a. Therefore, in this case, a = b = c. Wait, but earlier when I took a = b = 2, c = 1, which would mean t = c / (a + b) = 1 / 4, not 1/2. Wait, so there was a mistake here. Earlier, when I set t = y / x = c / x, where x = a + b. But in the substitution where I set t = y / x, then with a = b, x = 2a, y = c, so t = c / (2a). Therefore, when I found t = 1/2, that means c / (2a) = 1/2 => c = a. Therefore, in this case, a = b = c. Wait, but earlier when I took a = b = 2, c = 1, that would correspond to t = 1 / (2*2) = 1/4, not 1/2. Therefore, my previous example where I took a = b = 2, c = 1 was not corresponding to t = 1/2. Wait, but earlier when I calculated with a = b = 2, c = 1, I found the value to be 100. But according to the substitution, when t = 1/2, c = a. Therefore, if I take a = b = 2, c = 2, then t = 2 / (2 + 2) = 2 / 4 = 1/2. Let me compute the value in this case:

a = b = 2, c = 2.

Numerator: (2 + 2 + 2)[(2 + 2)^2 + (2 + 2 + 4*2)^2] = 6 * [16 + (8 + 8)^2] = 6 * [16 + 256] = 6 * 272 = 1632

Denominator: abc = 2 * 2 * 2 = 8

Value: 1632 / 8 = 204. Which is higher than 100. Wait, but earlier when I took a = b = 2, c = 1, I had value 100. How is that possible?

Wait, there must be confusion here. Let me re-examine the substitution. Earlier, when I set t = y / x, where x = a + b, y = c. Then, when I found t = 1/2, that corresponds to c = (1/2)x. Therefore, if x = a + b, then c = (a + b)/2. Therefore, if I set a = b, then x = 2a, so c = (2a)/2 = a. Therefore, c = a. Therefore, in this case, a = b = c. But when I took a = b = 2, c = 1, that would mean c = 1, which is not equal to a. Therefore, that case does not correspond to t = 1/2. Therefore, my previous calculation where I took a = b = 2, c = 1 was not under the assumption that t = 1/2. Therefore, there was a mistake in my previous reasoning. Wait, but earlier when I did the substitution with t = y / x, and found that t = 1/2 gives the minimum, but when I took a = b = 2, c = 1, which is t = 1 / (2 + 2) = 1/4, not 1/2. Therefore, how did I get 100 in that case?

Wait, let me recalculate that case. If a = b = 2, c = 1:

Numerator: (a + b + c) = 2 + 2 + 1 = 5

(a + b)^2 + (a + b + 4c)^2 = (4)^2 + (4 + 4*1)^2 = 16 + 8^2 = 16 + 64 = 80

Numerator total: 5 * 80 = 400

Denominator: abc = 2 * 2 * 1 = 4

Value: 400 / 4 = 100. Correct. But according to the substitution, when t = c / (a + b) = 1 / 4, which is different from t = 1/2. Therefore, how come this case gives the same value as the critical point? There must be some error in my substitution.

Wait, earlier when I did substitution with t = y / x, where x = a + b, y = c, then I set t = y / x, and found that the minimum occurs at t = 1/2. But in this case, with t = 1/4, I still get the value 100. Which suggests that my previous calculation of the critical point was wrong? Or there is some miscalculation.

Wait, let me go back. When I did substitution with t = y / x, then I had:

f(t) = 8*(2 + t)(1 + 2t + 2t²)/t. Then I found critical point at t = 1/2. But when I plug t = 1/4 into this function:

f(1/4) = 8*(2 + 1/4)*(1 + 2*(1/4) + 2*(1/4)^2) / (1/4)

Compute step by step:

First, 2 + 1/4 = 9/4

Next, 1 + 2*(1/4) + 2*(1/16) = 1 + 1/2 + 1/8 = (8/8 + 4/8 + 1/8) = 13/8

Therefore, numerator inside: 9/4 * 13/8 = 117/32

Then, f(t) = 8 * (117/32) / (1/4) = 8 * (117/32) * 4 = 8 * (117/8) = 117. Which is not 100. But earlier, when I took a = b = 2, c = 1, which corresponds to t = 1/4, I got 100. Therefore, there is a contradiction here. Which means there was a mistake in my substitution.

Wait, let me check my substitution again. Earlier, I set a = b = x, c = y. Then, I set t = y / x. Then, I derived the expression f(t) = 8*(2 + t)(1 + 2t + 2t²)/t. But when I took a = b = 2, c = 1, then t = y / x = 1 / 2 = 0.5. Wait, no! Wait, if a = b = x = 2, then x = 2, and c = y = 1. Therefore, t = y / x = 1 / 2 = 0.5. Therefore, t = 1/2. Therefore, in this case, t = 1/2, and f(t) = 100. Which matches. Earlier, I thought that t = 1/4, but that was a mistake. Therefore, my previous calculation where I thought t = 1/4 was wrong. Therefore, the case a = b = 2, c = 1 corresponds to t = 1/2, which is the critical point. Therefore, everything is consistent. Therefore, the minimum value is indeed 100.

Therefore, after careful checking, I can confirm that the minimum value is 100. Therefore, the answer is \boxed{100}.

**Final Answer**
\boxed{100}

To find the minimum value of the expression:

$$
\frac{(a + b + c)[(a + b)^2 + (a + b + 4c)^2]}{abc}
$$

where $ a, b, c > 0 $, we begin by analyzing the structure of the expression. It is symmetric in $ a $ and $ b $, which suggests that the minimum may occur when $ a = b $. This symmetry allows us to simplify the problem by setting $ a = b = x $ and $ c = y $, reducing the number of variables.

---

### Step 1: Substitution and Simplification

Let $ a = b = x $, $ c = y $. Then:

- $ a + b + c = 2x + y $
- $ (a + b)^2 = (2x)^2 = 4x^2 $
- $ (a + b + 4c)^2 = (2x + 4y)^2 = 4x^2 + 16xy + 16y^2 $

So the numerator becomes:

$$
(2x + y)\left[4x^2 + (2x + 4y)^2\right] = (2x + y)\left[4x^2 + 4x^2 + 16xy + 16y^2\right] = (2x + y)(8x^2 + 16xy + 16y^2)
$$

The denominator is:

$$
abc = x \cdot x \cdot y = x^2 y
$$

Thus, the expression becomes:

$$
\frac{(2x + y)(8x^2 + 16xy + 16y^2)}{x^2 y}
$$

---

### Step 2: Introduce a New Variable

Let $ t = \frac{y}{x} $, so $ y = tx $. Substituting into the expression:

- $ 2x + y = 2x + tx = x(2 + t) $
- $ 8x^2 + 16xy + 16y^2 = 8x^2 + 16x(tx) + 16(tx)^2 = x^2(8 + 16t + 16t^2) $

So the expression becomes:

$$
\frac{x(2 + t) \cdot x^2(8 + 16t + 16t^2)}{x^2 \cdot tx} = \frac{x^3(2 + t)(8 + 16t + 16t^2)}{x^3 t} = \frac{(2 + t)(8 + 16t + 16t^2)}{t}
$$

Factor the numerator:

$$
8 + 16t + 16t^2 = 8(1 + 2t + 2t^2)
$$

So the expression becomes:

$$
\frac{(2 + t) \cdot 8(1 + 2t + 2t^2)}{t} = 8 \cdot \frac{(2 + t)(1 + 2t + 2t^2)}{t}
$$

---

### Step 3: Minimize the Function

Let:

$$
f(t) = 8 \cdot \frac{(2 + t)(1 + 2t + 2t^2)}{t}
$$

We now minimize $ f(t) $ for $ t > 0 $. Expanding the numerator:

$$
(2 + t)(1 + 2t + 2t^2) = 2(1 + 2t + 2t^2) + t(1 + 2t + 2t^2) = 2 + 4t + 4t^2 + t + 2t^2 + 2t^3 = 2 + 5t + 6t^2 + 2t^3
$$

So:

$$
f(t) = 8 \cdot \frac{2 + 5t + 6t^2 + 2t^3}{t}
$$

Now, take the derivative of $ f(t) $ to find the critical points. Let:

$$
f(t) = 8 \left( \frac{2}{t} + 5 + 6t + 2t^2 \right)
$$

Then:

$$
f'(t) = 8 \left( -\frac{2}{t^2} + 6 + 4t \right)
$$

Set $ f'(t) = 0 $:

$$
-\frac{2}{t^2} + 6 + 4t = 0 \Rightarrow 4t^3 + 6t^2 - 2 = 0
$$

Solving this cubic equation, we find the positive root is:

$$
t = \frac{1}{2}
$$

---

### Step 4: Evaluate the Minimum

Substitute $ t = \frac{1}{2} $ into the expression:

$$
f\left(\frac{1}{2}\right) = 8 \cdot \frac{(2 + \frac{1}{2})(1 + 2 \cdot \frac{1}{2} + 2 \cdot \left(\frac{1}{2}\right)^2)}{\frac{1}{2}}
$$

Compute each term:

- $ 2 + \frac{1}{2} = \frac{5}{2} $
- $ 1 + 1 + \frac{1}{2} = \frac{5}{2} $
- So the numerator is $ \frac{5}{2} \cdot \frac{5}{2} = \frac{25}{4} $
- Denominator is $ \frac{1}{2} $

Thus:

$$
f\left(\frac{1}{2}\right) = 8 \cdot \frac{25/4}{1/2} = 8 \cdot \frac{25}{2} = 100
$$

---

### Step 5: Verify the Minimum

We can verify that this is indeed the minimum by checking the behavior of $ f(t) $ as $ t \to 0^+ $ and $ t \to \infty $, both of which cause $ f(t) \to \infty $. Therefore, the function has a global minimum at $ t = \frac{1}{2} $, and the minimum value is:

$$
\boxed{100}
$$

Solution:
By AM-GM,
\[a + b \ge 2 \sqrt{ab},\]so $(a + b)^2 \ge 4ab.$

Also by AM-GM,
\[(a + 2c) + (b + 2c) \ge 2 \sqrt{(a + 2c)(b + 2c)},\]so $(a + b + 4c)^2 \ge 4(a + 2c)(b + 2c).$

Hence,
\begin{align*}
(a + b)^2 + (a + b + 4c)^2 &\ge 4ab + 4(a + 2c)(b + 2c) \\
&= 8ab + 8ac + 8bc + 16c^2 \\
&= 8(ab + ac + bc + 2c^2).
\end{align*}By AM-GM,
\begin{align*}
ab + ac + bc + 2c^2 &= \frac{ab}{2} + \frac{ab}{2} + ac + bc + 2c^2 \\
&\ge 5 \sqrt[5]{\frac{ab}{2} \cdot \frac{ab}{2} \cdot ac \cdot bc \cdot 2c^2} \\
&= 5 \sqrt[5]{\frac{a^3 b^3 c^4}{2}}.
\end{align*}Also by AM-GM,
\begin{align*}
a + b + c &= \frac{a}{2} + \frac{a}{2} + \frac{b}{2} + \frac{b}{2} + c \\
&\ge 5 \sqrt[5]{\frac{a}{2} \cdot \frac{a}{2} \cdot \frac{b}{2} \cdot \frac{b}{2} \cdot c} \\
&= 5 \sqrt[5]{\frac{a^2 b^2 c}{16}}.
\end{align*}Hence,
\begin{align*}
\frac{(a + b + c)[(a + b)^2 + (a + b + 4c)^2]}{abc} &\ge 8 \cdot \frac{5 \sqrt[5]{\frac{a^2 b^2 c}{16}} \cdot 5 \sqrt[5]{\frac{a^3 b^3 c^4}{2}}}{abc} \\
&= 100.
\end{align*}Equality occurs when $a = b = 2$ and $c = 1,$ so the minimum value is $\boxed{100}.$


Explanation:
Okay, so I need to find all real numbers x such that -4 < x⁴ + 4x² < 21. Hmm, let me think about how to approach this.

First, the inequality is a compound inequality, meaning I have to solve both parts separately and then find the intersection of the solutions. Let me write that down:

-4 < x⁴ + 4x² and x⁴ + 4x² < 21.

Starting with the first inequality: -4 < x⁴ + 4x².

Wait, x⁴ and x² are both even powers, so they are always non-negative, right? That means x⁴ + 4x² is always greater than or equal to zero. But the left side of the inequality is -4, which is negative. So if the left side is always non-negative, then -4 < something non-negative is always true. Therefore, the first inequality is always satisfied for all real numbers x.

So, the first part doesn't restrict the values of x. Therefore, I only need to focus on the second inequality: x⁴ + 4x² < 21.

Alright, so let me work on solving x⁴ + 4x² < 21.

Maybe I can rewrite this inequality to make it easier to solve. Let me set y = x². Since x² is always non-negative, y ≥ 0. Then the inequality becomes:

y² + 4y < 21.

Let me rearrange this to standard quadratic form:

y² + 4y - 21 < 0.

Now, solving the quadratic inequality y² + 4y - 21 < 0. To do this, I need to find the roots of the equation y² + 4y - 21 = 0.

Using the quadratic formula: y = [-b ± √(b² - 4ac)] / 2a. Here, a = 1, b = 4, c = -21.

Calculating discriminant D: D = 16 - 4*1*(-21) = 16 + 84 = 100.

So, the roots are y = [-4 ± √100]/2 = [-4 ± 10]/2.

Calculating both roots:

First root: (-4 + 10)/2 = 6/2 = 3.

Second root: (-4 - 10)/2 = -14/2 = -7.

So, the quadratic equation factors as (y - 3)(y + 7) = 0? Wait, let me check. If the roots are 3 and -7, then the quadratic is (y - 3)(y + 7) = y² + 7y - 3y -21 = y² + 4y -21. Yes, that's correct.

So, the inequality y² + 4y -21 < 0 is equivalent to (y - 3)(y + 7) < 0.

To solve this inequality, I need to determine where the product is negative. The critical points are at y = -7 and y = 3. Since y = x² ≥ 0, we can ignore y = -7 because y cannot be negative.

So, considering y ≥ 0, the critical point is at y = 3. Let me analyze the intervals:

For y < -7: Not possible since y ≥ 0.

For -7 < y < 3: Since y ≥ 0, this interval becomes 0 ≤ y < 3.

For y > 3: y > 3.

Now, the quadratic (y - 3)(y + 7) is a parabola opening upwards (since coefficient of y² is positive). Therefore, it is negative between its roots. But since y cannot be less than -7, the inequality (y - 3)(y + 7) < 0 holds for y between -7 and 3. However, since y ≥ 0, the valid interval is 0 ≤ y < 3.

Therefore, the solution for y is 0 ≤ y < 3. Remember that y = x², so substituting back:

0 ≤ x² < 3.

Solving for x:

x² < 3 implies that -√3 < x < √3. Also, since x² ≥ 0, which is always true. Therefore, combining these, the solution is x ∈ (-√3, √3).

Wait, but let me check if I did everything correctly. Let me verify with some test values.

Take x = 0: x⁴ + 4x² = 0 + 0 = 0 < 21. True.

Take x = 2: x⁴ + 4x² = 16 + 16 = 32. Which is greater than 21. So, x = 2 is not in the solution. But √3 is approximately 1.732, so 2 is outside the interval. So that seems okay.

Take x = 1: 1 + 4 = 5 < 21. True.

Take x approaching √3: Let me compute x⁴ + 4x² when x = √3. Let me compute:

x² = 3, so x⁴ = (x²)² = 9. Therefore, x⁴ + 4x² = 9 + 12 = 21. So, at x = ±√3, the expression equals 21. Since the inequality is strict (< 21), the endpoints are excluded. Therefore, the solution is indeed -√3 < x < √3.

But wait, the original problem was -4 < x⁴ + 4x² < 21. Since we already established that the left inequality is always true, the entire solution set is just the solution to the right inequality, which is (-√3, √3).

But let me check if there are any other constraints. For example, could there be values of x where x⁴ + 4x² is less than -4? But since x⁴ and 4x² are both non-negative, their sum is non-negative, so it can't be less than -4. Therefore, the left inequality is always true, so the entire solution is just the interval where x⁴ + 4x² < 21, which is (-√3, √3).

But let me check if there's a mistake here. Let me think again. Suppose someone might think that maybe for some x, x⁴ + 4x² could be negative? But since x⁴ is non-negative and 4x² is non-negative, their sum is non-negative. Therefore, the left inequality is always true. Therefore, the answer is indeed (-√3, √3).

But let me check if I made a mistake in solving the quadratic inequality. Let me go through that again.

Original inequality after substitution: y² + 4y -21 < 0. The roots are y = 3 and y = -7. The quadratic opens upwards, so it's negative between -7 and 3. Since y = x² ≥ 0, the valid interval is 0 ≤ y < 3. Therefore, x² < 3, so |x| < √3. Therefore, x ∈ (-√3, √3).

Yes, that seems correct.

But let me check with another value. Let me take x = √3. Then x⁴ + 4x² = ( (√3)^4 ) + 4*(√3)^2.

(√3)^2 = 3, so (√3)^4 = (3)^2 = 9. Then 4*(√3)^2 = 4*3 = 12. So total is 9 + 12 = 21. Which is equal to 21, so not less than 21. Therefore, x = √3 is excluded. Similarly for x = -√3.

Therefore, the interval is open.

Therefore, the answer is (-√3, √3).

But let me check if there's a possibility that maybe the original inequality could have more solutions? For example, maybe if x is very large, but since x⁴ dominates, as |x| increases, x⁴ + 4x² will go to infinity, so it will be greater than 21. Therefore, only between -√3 and √3 is the expression less than 21.

Therefore, combining everything, the solution is all real numbers between -√3 and √3.

But let me check if there's a mistake in substitution. Let me try solving the original inequality without substitution.

Original inequality: x⁴ + 4x² < 21.

Let me write it as x⁴ + 4x² - 21 < 0.

Let me try to factor this quartic equation. Maybe factor it as a quadratic in x².

So, x⁴ + 4x² - 21 = (x²)^2 + 4x² - 21. Which is the same as before. So, we can factor it as (x² + a)(x² + b) = x⁴ + (a + b)x² + ab. Comparing coefficients:

a + b = 4

ab = -21

So solving for a and b. The numbers that multiply to -21 and add to 4. Let me think. Factors of -21: 7 and -3. 7 + (-3) = 4. Yes! So, a = 7, b = -3. Therefore, the quartic factors as (x² + 7)(x² - 3).

Therefore, x⁴ + 4x² - 21 = (x² + 7)(x² - 3).

Therefore, the inequality becomes (x² + 7)(x² - 3) < 0.

Now, since x² + 7 is always positive (because x² ≥ 0, so x² + 7 ≥ 7 > 0), the sign of the product depends on (x² - 3). Therefore, the inequality reduces to (x² - 3) < 0, because the other factor is always positive.

Therefore, x² - 3 < 0 => x² < 3 => |x| < √3.

Which is the same result as before. Therefore, the solution is indeed -√3 < x < √3.

Therefore, combining with the first inequality which is always true, the final answer is (-√3, √3).

But let me check once more with another approach. Let me graph the function f(x) = x⁴ + 4x². Since it's an even function, symmetric about the y-axis. The minimum value occurs at x = 0, where f(0) = 0. As |x| increases, the function increases to infinity. Therefore, the graph is a U-shaped curve with the minimum at (0, 0).

So, the inequality -4 < f(x) < 21. Since f(x) is always ≥ 0, the left inequality is always true. The right inequality is f(x) < 21. Since the function increases as |x| increases, the solutions to f(x) < 21 are the x-values between -√3 and √3, as we found.

Therefore, I think this confirms the answer.

But just to be thorough, let me check if there are any other points where f(x) could be less than 21. For example, suppose x is very large, say x = 10. Then f(10) = 10^4 + 4*(10)^2 = 10000 + 400 = 10400, which is way larger than 21. So, definitely, beyond √3, the function is larger than 21.

What about between -√3 and √3? Let me take x = 1.5, which is less than √3 ≈ 1.732. Then f(1.5) = (1.5)^4 + 4*(1.5)^2.

Calculating:

(1.5)^2 = 2.25

(1.5)^4 = (2.25)^2 = 5.0625

Then 4*(2.25) = 9.

So total is 5.0625 + 9 = 14.0625 < 21. So yes, it's less than 21.

If I take x approaching √3, say x = 1.7, then x² = 2.89, x⁴ = (2.89)^2 ≈ 8.3521. Then 4x² ≈ 11.56. Total ≈ 8.3521 + 11.56 ≈ 19.9121 < 21. So still less than 21.

At x = √3, it's exactly 21. So, the interval is correct.

Therefore, I think I can be confident that the solution is all real numbers between -√3 and √3.

Therefore, in interval notation, that is (-√3, √3).

But let me check if the question requires any specific formatting or if there's a possibility of other solutions.

Wait, the original problem says "real values of x" so we don't have to worry about complex numbers. Also, since we have already considered all possibilities, and the first inequality is always true, the answer is indeed (-√3, √3).

Therefore, I think that's the final answer.

**Final Answer**
The solution in interval notation is \boxed{(-\sqrt{3}, \sqrt{3})}.

To solve the inequality
$$
-4 < x^4 + 4x^2 < 21,
$$
we analyze it in two parts:

---

### **Step 1: Analyze the Left Inequality**

$$
-4 < x^4 + 4x^2
$$

Note that both $ x^4 $ and $ 4x^2 $ are **non-negative** for all real $ x $, so their sum $ x^4 + 4x^2 \geq 0 $. Since $ -4 < 0 $, the inequality
$$
-4 < x^4 + 4x^2
$$
is **always true** for all real $ x $. Therefore, this inequality imposes **no restriction** on the values of $ x $.

---

### **Step 2: Analyze the Right Inequality**

$$
x^4 + 4x^2 < 21
$$

Let’s simplify this by substituting $ y = x^2 $, where $ y \geq 0 $. Then the inequality becomes:

$$
y^2 + 4y < 21
$$

Rewriting in standard quadratic form:

$$
y^2 + 4y - 21 < 0
$$

We solve the corresponding equation:

$$
y^2 + 4y - 21 = 0
$$

Using the quadratic formula:

$$
y = \frac{-4 \pm \sqrt{16 + 84}}{2} = \frac{-4 \pm \sqrt{100}}{2} = \frac{-4 \pm 10}{2}
$$

So the roots are:

$$
y = 3 \quad \text{and} \quad y = -7
$$

Since $ y = x^2 \geq 0 $, we discard $ y = -7 $, and consider the interval:

$$
0 \leq y < 3
$$

Substituting back $ y = x^2 $, we get:

$$
0 \leq x^2 < 3
$$

Taking square roots:

$$
- \sqrt{3} < x < \sqrt{3}
$$

---

### **Step 3: Combine the Results**

From Step 1, the left inequality is always true.
From Step 2, the right inequality restricts $ x $ to the interval $ (-\sqrt{3}, \sqrt{3}) $.

Thus, the solution to the original compound inequality is:

$$
\boxed{(-\sqrt{3}, \sqrt{3})}
$$

[Assistant]
Explanation:
Okay, so I need to find all real numbers x such that -4 < x⁴ + 4x² < 21. Hmm, let me think about how to approach this.

First, the inequality is a compound inequality, meaning I have to solve both parts separately and then find the intersection of the solutions. Let me write that down:

-4 < x⁴ + 4x² and x⁴ + 4x² < 21.

Starting with the first inequality: -4 < x⁴ + 4x².

Wait, x⁴ and x² are both even powers, so they are always non-negative, right? That means x⁴ + 4x² is always greater than or equal to zero. But the left side of the inequality is -4, which is negative. So if the left side is always non-negative, then -4 < something non-negative is always true. Therefore, the first inequality is always satisfied for all real numbers x.

So, the first part doesn't restrict the values of x. Therefore, I only need to focus on the second inequality: x⁴ + 4x² < 21.

Alright, so let me work on solving x⁴ + 4x² < 21.

Maybe I can rewrite this inequality to make it easier to solve. Let me set y = x². Since x² is always non-negative, y ≥ 0. Then the inequality becomes:

y² + 4y < 21.

Let me rearrange this to standard quadratic form:

y² + 4y - 21 < 0.

Now, solving the quadratic inequality y² + 4y - 21 < 0. To do this, I need to find the roots of the equation y² + 4y - 21 = 0.

Using the quadratic formula: y = [-b ± √(b² - 4ac)] / 2a. Here, a = 1, b = 4, c = -21.

Calculating discriminant D: D = 16 - 4*1*(-21) = 16 + 84 = 100.

So, the roots are y = [-4 ± √100]/2 = [-4 ± 10]/2.

Calculating both roots:

First root: (-4 + 10)/2 = 6/2 = 3.

Second root: (-4 - 10)/2 = -14/2 = -7.

So, the quadratic equation factors as (y - 3)(y + 7) = 0? Wait, let me check. If the roots are 3 and -7, then the quadratic is (y - 3)(y + 7) = y² + 7y - 3y -21 = y² + 4y -21. Yes, that's correct.

So, the inequality y² + 4y -21 < 0 is equivalent to (y - 3)(y + 7) < 0.

To solve this inequality, I need to determine where the product is negative. The critical points are at y = -7 and y = 3. Since y = x² ≥ 0, we can ignore y = -7 because y cannot be negative.

So, considering y ≥ 0, the critical point is at y = 3. Let me analyze the intervals:

For y < -7: Not possible since y ≥ 0.

For -7 < y < 3: Since y ≥ 0, this interval becomes 0 ≤ y < 3.

For y > 3: y > 3.

Now, the quadratic (y - 3)(y + 7) is a parabola opening upwards (since coefficient of y² is positive). Therefore, it is negative between its roots. But since y cannot be less than -7, the inequality (y - 3)(y + 7) < 0 holds for y between -7 and 3. However, since y ≥ 0, the valid interval is 0 ≤ y < 3.

Therefore, the solution for y is 0 ≤ y < 3. Remember that y = x², so substituting back:

0 ≤ x² < 3.

Solving for x:

x² < 3 implies that -√3 < x < √3. Also, since x² ≥ 0, which is always true. Therefore, combining these, the solution is x ∈ (-√3, √3).

Wait, but let me check if I did everything correctly. Let me verify with some test values.

Take x = 0: x⁴ + 4x² = 0 + 0 = 0 < 21. True.

Take x = 2: x⁴ + 4x² = 16 + 16 = 32. Which is greater than 21. So, x = 2 is not in the solution. But √3 is approximately 1.732, so 2 is outside the interval. So that seems okay.

Take x = 1: 1 + 4 = 5 < 21. True.

Take x approaching √3: Let me compute x⁴ + 4x² when x = √3. Let me compute:

x² = 3, so x⁴ = (x²)² = 9. Therefore, x⁴ + 4x² = 9 + 12 = 21. So, at x = ±√3, the expression equals 21. Since the inequality is strict (< 21), the endpoints are excluded. Therefore, the solution is indeed -√3 < x < √3.

But wait, the original problem was -4 < x⁴ + 4x² < 21. Since we already established that the left inequality is always true, the entire solution set is just the solution to the right inequality, which is (-√3, √3).

But let me check if there are any other constraints. For example, could there be values of x where x⁴ + 4x² is less than -4? But since x⁴ and 4x² are both non-negative, their sum is non-negative, so it can't be less than -4. Therefore, the left inequality is always true, so the entire solution is just the interval where x⁴ + 4x² < 21, which is (-√3, √3).

But let me check if there's a mistake here. Let me think again. Suppose someone might think that maybe for some x, x⁴ + 4x² could be negative? But since x⁴ is non-negative and 4x² is non-negative, their sum is non-negative. Therefore, the left inequality is always true. Therefore, the answer is indeed (-√3, √3).

But let me check if I made a mistake in solving the quadratic inequality. Let me go through that again.

Original inequality after substitution: y² + 4y -21 < 0. The roots are y = 3 and y = -7. The quadratic opens upwards, so it's negative between -7 and 3. Since y = x² ≥ 0, the valid interval is 0 ≤ y < 3. Therefore, x² < 3, so |x| < √3. Therefore, x ∈ (-√3, √3).

Yes, that seems correct.

But let me check with another value. Let me take x = √3. Then x⁴ + 4x² = ( (√3)^4 ) + 4*(√3)^2.

(√3)^2 = 3, so (√3)^4 = (3)^2 = 9. Then 4*(√3)^2 = 4*3 = 12. So total is 9 + 12 = 21. Which is equal to 21, so not less than 21. Therefore, x = √3 is excluded. Similarly for x = -√3.

Therefore, the interval is open.

Therefore, the answer is (-√3, √3).

But let me check if there's a possibility that maybe the original inequality could have more solutions? For example, maybe if x is very large, but since x⁴ dominates, as |x| increases, x⁴ + 4x² will go to infinity, so it will be greater than 21. Therefore, only between -√3 and √3 is the expression less than 21.

Therefore, combining everything, the solution is all real numbers between -√3 and √3.

But let me check if there's a mistake in substitution. Let me try solving the original inequality without substitution.

Original inequality: x⁴ + 4x² < 21.

Let me write it as x⁴ + 4x² - 21 < 0.

Let me try to factor this quartic equation. Maybe factor it as a quadratic in x².

So, x⁴ + 4x² - 21 = (x²)^2 + 4x² - 21. Which is the same as before. So, we can factor it as (x² + a)(x² + b) = x⁴ + (a + b)x² + ab. Comparing coefficients:

a + b = 4

ab = -21

So solving for a and b. The numbers that multiply to -21 and add to 4. Let me think. Factors of -21: 7 and -3. 7 + (-3) = 4. Yes! So, a = 7, b = -3. Therefore, the quartic factors as (x² + 7)(x² - 3).

Therefore, x⁴ + 4x² - 21 = (x² + 7)(x² - 3).

Therefore, the inequality becomes (x² + 7)(x² - 3) < 0.

Now, since x² + 7 is always positive (because x² ≥ 0, so x² + 7 ≥ 7 > 0), the sign of the product depends on (x² - 3). Therefore, the inequality reduces to (x² - 3) < 0, because the other factor is always positive.

Therefore, x² - 3 < 0 => x² < 3 => |x| < √3.

Which is the same result as before. Therefore, the solution is indeed -√3 < x < √3.

Therefore, combining with the first inequality which is always true, the final answer is (-√3, √3).

But let me check once more with another approach. Let me graph the function f(x) = x⁴ + 4x². Since it's an even function, symmetric about the y-axis. The minimum value occurs at x = 0, where f(0) = 0. As |x| increases, the function increases to infinity. Therefore, the graph is a U-shaped curve with the minimum at (0, 0).

So, the inequality -4 < f(x) < 21. Since f(x) is always ≥ 0, the left inequality is always true. The right inequality is f(x) < 21. Since the function increases as |x| increases, the solutions to f(x) < 21 are the x-values between -√3 and √3, as we found.

Therefore, I think this confirms the answer.

But just to be thorough, let me check if there are any other points where f(x) could be less than 21. For example, suppose x is very large, say x = 10. Then f(10) = 10^4 + 4*(10)^2 = 10000 + 400 = 10400, which is way larger than 21. So, definitely, beyond √3, the function is larger than 21.

What about between -√3 and √3? Let me take x = 1.5, which is less than √3 ≈ 1.732. Then f(1.5) = (1.5)^4 + 4*(1.5)^2.

Calculating:

(1.5)^2 = 2.25

(1.5)^4 = (2.25)^2 = 5.0625

Then 4*(2.25) = 9.

So total is 5.0625 + 9 = 14.0625 < 21. So yes, it's less than 21.

If I take x approaching √3, say x = 1.7, then x² = 2.89, x⁴ = (2.89)^2 ≈ 8.3521. Then 4x² ≈ 11.56. Total ≈ 8.3521 + 11.56 ≈ 19.9121 < 21. So still less than 21.

At x = √3, it's exactly 21. So, the interval is correct.

Therefore, I think I can be confident that the solution is all real numbers between -√3 and √3.

Therefore, in interval notation, that is (-√3, √3).

But let me check if the question requires any specific formatting or if there's a possibility of other solutions.

Wait, the original problem says "real values of x" so we don't have to worry about complex numbers. Also, since we have already considered all possibilities, and the first inequality is always true, the answer is indeed (-√3, √3).

Therefore, I think that's the final answer.

**Final Answer**
The solution in interval notation is \boxed{(-\sqrt{3}, \sqrt{3})}.

To solve the inequality
$$
-4 < x^4 + 4x^2 < 21,
$$
we analyze it in two parts:

---

### **Step 1: Analyze the Left Inequality**

$$
-4 < x^4 + 4x^2
$$

Note that both $ x^4 $ and $ 4x^2 $ are **non-negative** for all real $ x $, so their sum $ x^4 + 4x^2 \geq 0 $. Since $ -4 < 0 $, the inequality
$$
-4 < x^4 + 4x^2
$$
is **always true** for all real $ x $. Therefore, this inequality imposes **no restriction** on the values of $ x $.

---

### **Step 2: Analyze the Right Inequality**

$$
x^4 + 4x^2 < 21
$$

Let’s simplify this by substituting $ y = x^2 $, where $ y \geq 0 $. Then the inequality becomes:

$$
y^2 + 4y < 21
$$

Rewriting in standard quadratic form:

$$
y^2 + 4y - 21 < 0
$$

We solve the corresponding equation:

$$
y^2 + 4y - 21 = 0
$$

Using the quadratic formula:

$$
y = \frac{-4 \pm \sqrt{16 + 84}}{2} = \frac{-4 \pm \sqrt{100}}{2} = \frac{-4 \pm 10}{2}
$$

So the roots are:

$$
y = 3 \quad \text{and} \quad y = -7
$$

Since $ y = x^2 \geq 0 $, we discard $ y = -7 $, and consider the interval:

$$
0 \leq y < 3
$$

Substituting back $ y = x^2 $, we get:

$$
0 \leq x^2 < 3
$$

Taking square roots:

$$
- \sqrt{3} < x < \sqrt{3}
$$

---

### **Step 3: Combine the Results**

From Step 1, the left inequality is always true.
From Step 2, the right inequality restricts $ x $ to the interval $ (-\sqrt{3}, \sqrt{3}) $.

Thus, the solution to the original compound inequality is:

$$
\boxed{(-\sqrt{3}, \sqrt{3})}
$$

Solution:
Let us first define $y=x^{2}$. We can then plug this value into the inequality and add 4 to $-4$, $x^4+4x^2$, and 21, yielding $$0<y^{2}+4y+4<25.$$We can factor $y^2+4y+4$ to obtain $$0<(y+2)^{2}<25.$$Taking the square root gives us $0<|y+2|<5$, which in turn gives us two intervals for solutions of $y$: $-2<y<3$, or $-7<y<-2$.

However, $y$ must be non-negative since $y=x^{2}$, so we have $0\leq y<3$. This means that $-\sqrt{3}< x<\sqrt{3}$ satisfies the original inequality. In interval notation, this is $\boxed{(-\sqrt{3}, \sqrt{3})}$.


Explanation:
Okay, so I need to figure out how many possible values of x there are when dealing a 54-card deck to x players, with each player getting y cards. The constraints are that x is at least 2 and y is at least 5. Let me start by understanding the problem step by step.

First, the deck has 54 cards. When dealing all the cards to x players, each player gets y cards. So, the total number of cards dealt should be equal to the number of players multiplied by the number of cards each player gets. That gives me the equation:

x * y = 54

So, x and y are positive integers that satisfy this equation. But there are constraints: x ≥ 2 and y ≥ 5. Therefore, I need to find all possible integer values of x (number of players) such that when you divide 54 by x, you get an integer y that is at least 5.

Alternatively, since x and y are positive integers, x must be a divisor of 54. So, first, I should find all the divisors of 54, then check which ones satisfy the conditions on x and y.

Let me recall that to find the divisors of 54, I can start by prime factorizing 54. Let me do that.

54 can be divided by 2 first: 54 ÷ 2 = 27. Then 27 is 3³. So, the prime factorization of 54 is 2 * 3³. Therefore, the divisors of 54 can be found by taking all combinations of exponents for the primes 2 and 3.

The exponents for 2 can be 0 or 1 (since 2^1 is part of the factorization, and 2^0 is also allowed). The exponents for 3 can be 0, 1, 2, or 3. Therefore, the total number of divisors is (1+1)*(3+1) = 2*4=8. Let me list them:

For 2^0 * 3^0 = 1

2^0 * 3^1 = 3

2^0 * 3^2 = 9

2^0 * 3^3 = 27

2^1 * 3^0 = 2

2^1 * 3^1 = 6

2^1 * 3^2 = 18

2^1 * 3^3 = 54

So the divisors of 54 are: 1, 2, 3, 6, 9, 18, 27, 54.

But remember, x has to be at least 2. So we can eliminate 1. So possible x values are 2, 3, 6, 9, 18, 27, 54. However, we also have the constraint that y must be at least 5. Since y = 54 / x, we need 54 / x ≥ 5. Let me write that inequality:

54 / x ≥ 5

Solving for x:

Multiply both sides by x (assuming x positive, which it is):

54 ≥ 5x

Then divide both sides by 5:

54 / 5 ≥ x

Which is 10.8 ≥ x. Since x must be an integer, x ≤ 10.

Therefore, x must be a divisor of 54 that is at least 2 and at most 10. Let me check the list of possible x values (excluding 1): 2, 3, 6, 9, 18, 27, 54. Now, which of these are ≤ 10?

Looking at the list:

2, 3, 6, 9 are all ≤ 10. The next ones are 18, 27, 54, which are greater than 10. Therefore, the possible x values are 2, 3, 6, 9. Wait, but let me check if for each of these x values, y is indeed at least 5.

Let me verify:

For x=2: y=54/2=27 ≥5 ✔

x=3: y=54/3=18 ≥5 ✔

x=6: y=54/6=9 ≥5 ✔

x=9: y=54/9=6 ≥5 ✔

x=18: y=54/18=3, which is less than 5. So this is invalid.

Similarly, x=27: y=2, invalid. x=54: y=1, invalid. So indeed, only x=2,3,6,9 satisfy both x≥2 and y≥5.

Wait, but hold on. Let me check if there are other divisors of 54 between 2 and 10 that I might have missed. The divisors of 54 are 1,2,3,6,9,18,27,54. So between 2 and 10, we have 2,3,6,9. So that's four values. Therefore, the number of possible x is 4?

But wait, let me think again. Is there a possibility that x could be something else? For example, maybe if x is not a divisor of 54? Wait, but if x is not a divisor of 54, then y would not be an integer. But the problem says "I deal all of the cards to x players, with each player getting y cards." So, that implies that y must be an integer. Therefore, x must divide 54. Therefore, only divisors of 54 are possible. Therefore, the possible x's are only those divisors of 54 that satisfy x ≥2 and y=54/x ≥5. Which gives us x=2,3,6,9. Therefore, 4 possible values. Therefore, the answer is 4?

Wait, but let me check again. Let me think if there's another way to approach this. Maybe I made a mistake in the inequality?

Original equation: x*y=54, with x ≥2, y ≥5. So, x must be a divisor of 54, and also y=54/x ≥5. So, 54/x ≥5 => x ≤54/5=10.8. Since x is integer, x ≤10. So x must be a divisor of 54 between 2 and 10 inclusive. Divisors of 54 in that range are 2,3,6,9. So that's four values. Therefore, answer is 4? But let me check again if there are more divisors. For example, 54 divided by 2 is 27, which is okay. Divided by 3 is 18, okay. Divided by 6 is 9, okay. Divided by 9 is 6, okay. Divided by 10? Wait, 10 is not a divisor of 54. 54 divided by 10 is 5.4, which is not integer. So, no. So, yes, only 2,3,6,9. Therefore, 4 possible values. Therefore, the answer is 4?

But wait, hold on. Let me check if there's another way. Maybe considering that y must be at least 5, so for each x, y=54/x must be integer and ≥5. So, equivalently, x must be a divisor of 54, and x ≤54/5=10.8. So x can be 2,3,6,9. So 4 values. Therefore, answer is 4. But let me check again if there are other divisors between 2 and 10. Let me list all divisors again:

Divisors of 54: 1, 2, 3, 6, 9, 18, 27, 54. So between 2 and 10, inclusive, we have 2,3,6,9. So that's four numbers. Therefore, answer is 4. Therefore, the number of possible values of x is 4. Therefore, the answer is \boxed{4}?

Wait, but wait. Let me check if there's a mistake here. For example, if x=4, then y=54/4=13.5, which is not integer. Similarly, x=5: 54/5=10.8, not integer. x=7: 54/7≈7.71, not integer. x=8: 54/8=6.75, not integer. x=10: 54/10=5.4, not integer. So none of these are integers. Therefore, only the divisors. Therefore, yes, only 2,3,6,9. Therefore, 4 values. Therefore, answer is 4. But let me check again if the problem says "each player getting y cards", so y must be integer. Therefore, x must divide 54. Therefore, yes. Therefore, answer is 4. But wait, let me check once more.

Alternatively, maybe I need to consider that y must be at least 5, so for each x, y=54/x ≥5. Therefore, x ≤54/5=10.8. So x can be up to 10. So x can be 2,3,6,9. So 4 values. Therefore, answer is 4. Therefore, I think that is correct.

But let me check if there's a different interpretation. For example, maybe the problem allows for some cards to be left over? But the problem says "I deal all of the cards to x players", so all cards are dealt, so y must be integer. Therefore, x must divide 54. Therefore, the answer is 4. Therefore, I think the answer is 4. But let me check once again.

Wait, but maybe I made a mistake in the divisors. Let me list all divisors again. 54 divided by 1 is 54. Divided by 2 is 27. Divided by 3 is 18. Divided by 6 is 9. Divided by 9 is 6. Divided by 18 is 3. Divided by 27 is 2. Divided by 54 is 1. So divisors are 1,2,3,6,9,18,27,54. So yes, that's correct. So between 2 and 10, the divisors are 2,3,6,9. So four numbers. Therefore, answer is 4. Therefore, I think that is correct.

But wait, hold on. Let me check if x=1 is allowed? But the problem says x is at least 2, so x=1 is excluded. Therefore, yes. Therefore, answer is 4. Therefore, I think the answer is 4. Therefore, the number of possible values of x is 4. Therefore, the answer is \boxed{4}.

Wait, but hold on. Let me check once more. Maybe I need to check if y is at least 5. For x=2, y=27, which is okay. For x=3, y=18, okay. For x=6, y=9, okay. For x=9, y=6, okay. So all of these satisfy y ≥5. If x=1, y=54, but x must be at least 2. If x=18, y=3, which is less than 5. So invalid. Similarly for higher x. Therefore, yes, only 4 values. Therefore, answer is 4. Therefore, I think that is correct.

But wait, hold on. Let me check if there's another way to think about this. For example, if we think of possible values of y. Since y must be at least 5, and x must be at least 2, then y can be from 5 up to 54/2=27. But since x must be integer, y must divide 54. Wait, no. Wait, if x=54/y, so y must divide 54. Therefore, y must be a divisor of 54, and y ≥5. So, let me think. If I approach it from the y side, then y must be a divisor of 54, y ≥5, and x=54/y must be integer and ≥2. Therefore, same thing. So, divisors of 54 that are ≥5. Let me list them:

Divisors of 54: 1,2,3,6,9,18,27,54. So divisors ≥5 are 6,9,18,27,54. Then, for each of these y's, check if x=54/y is ≥2.

For y=6: x=54/6=9 ≥2 ✔

y=9: x=54/9=6 ≥2 ✔

y=18: x=54/18=3 ≥2 ✔

y=27: x=54/27=2 ≥2 ✔

y=54: x=54/54=1, which is less than 2. So invalid.

Therefore, the valid y's are 6,9,18,27. Therefore, corresponding x's are 9,6,3,2. Which is the same as before. Therefore, the number of x's is 4. Therefore, same answer. Therefore, answer is 4. Therefore, I think that is correct.

Therefore, I think the answer is 4. Therefore, the number of possible values of x is 4. Therefore, the answer is \boxed{4}.

**Final Answer**
\boxed{4}

To determine how many possible values of $ x $ (the number of players) are valid under the given constraints, we begin by analyzing the relationship between the number of players and the number of cards each player receives.

---

### Step 1: Understand the Relationship

We are told that:

- There are **54 cards** in the deck.
- All cards are dealt to $ x $ players.
- Each player receives $ y $ cards.
- Therefore, the total number of cards is:

$$
x \cdot y = 54
$$

This implies that $ x $ must be a **divisor of 54**, and $ y = \frac{54}{x} $ must be an integer.

---

### Step 2: Apply the Constraints

We are given two constraints:

- $ x \geq 2 $
- $ y \geq 5 $

From the equation $ y = \frac{54}{x} $, the second constraint becomes:

$$
\frac{54}{x} \geq 5 \Rightarrow x \leq \frac{54}{5} = 10.8
$$

Since $ x $ must be an integer, we conclude:

$$
2 \leq x \leq 10
$$

Now, we find all **divisors of 54** that lie in this range.

---

### Step 3: List the Divisors of 54

The prime factorization of 54 is:

$$
54 = 2 \cdot 3^3
$$

From this, we can list all the divisors of 54:

$$
1, 2, 3, 6, 9, 18, 27, 54
$$

Now, we filter these to find the ones that satisfy:

- $ x \geq 2 $
- $ x \leq 10 $

This gives us:

$$
x = 2, 3, 6, 9
$$

---

### Step 4: Verify the Corresponding $ y $ Values

For each of these $ x $ values, we compute $ y = \frac{54}{x} $:

- $ x = 2 \Rightarrow y = 27 $
- $ x = 3 \Rightarrow y = 18 $
- $ x = 6 \Rightarrow y = 9 $
- $ x = 9 \Rightarrow y = 6 $

All of these values of $ y $ are **greater than or equal to 5**, satisfying the constraint.

---

### Step 5: Final Count

We have found **4 valid values** of $ x $ that satisfy all the conditions:

$$
x = 2, 3, 6, 9
$$

---

### Final Answer

$$
\boxed{4}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many possible values of x there are when dealing a 54-card deck to x players, with each player getting y cards. The constraints are that x is at least 2 and y is at least 5. Let me start by understanding the problem step by step.

First, the deck has 54 cards. When dealing all the cards to x players, each player gets y cards. So, the total number of cards dealt should be equal to the number of players multiplied by the number of cards each player gets. That gives me the equation:

x * y = 54

So, x and y are positive integers that satisfy this equation. But there are constraints: x ≥ 2 and y ≥ 5. Therefore, I need to find all possible integer values of x (number of players) such that when you divide 54 by x, you get an integer y that is at least 5.

Alternatively, since x and y are positive integers, x must be a divisor of 54. So, first, I should find all the divisors of 54, then check which ones satisfy the conditions on x and y.

Let me recall that to find the divisors of 54, I can start by prime factorizing 54. Let me do that.

54 can be divided by 2 first: 54 ÷ 2 = 27. Then 27 is 3³. So, the prime factorization of 54 is 2 * 3³. Therefore, the divisors of 54 can be found by taking all combinations of exponents for the primes 2 and 3.

The exponents for 2 can be 0 or 1 (since 2^1 is part of the factorization, and 2^0 is also allowed). The exponents for 3 can be 0, 1, 2, or 3. Therefore, the total number of divisors is (1+1)*(3+1) = 2*4=8. Let me list them:

For 2^0 * 3^0 = 1

2^0 * 3^1 = 3

2^0 * 3^2 = 9

2^0 * 3^3 = 27

2^1 * 3^0 = 2

2^1 * 3^1 = 6

2^1 * 3^2 = 18

2^1 * 3^3 = 54

So the divisors of 54 are: 1, 2, 3, 6, 9, 18, 27, 54.

But remember, x has to be at least 2. So we can eliminate 1. So possible x values are 2, 3, 6, 9, 18, 27, 54. However, we also have the constraint that y must be at least 5. Since y = 54 / x, we need 54 / x ≥ 5. Let me write that inequality:

54 / x ≥ 5

Solving for x:

Multiply both sides by x (assuming x positive, which it is):

54 ≥ 5x

Then divide both sides by 5:

54 / 5 ≥ x

Which is 10.8 ≥ x. Since x must be an integer, x ≤ 10.

Therefore, x must be a divisor of 54 that is at least 2 and at most 10. Let me check the list of possible x values (excluding 1): 2, 3, 6, 9, 18, 27, 54. Now, which of these are ≤ 10?

Looking at the list:

2, 3, 6, 9 are all ≤ 10. The next ones are 18, 27, 54, which are greater than 10. Therefore, the possible x values are 2, 3, 6, 9. Wait, but let me check if for each of these x values, y is indeed at least 5.

Let me verify:

For x=2: y=54/2=27 ≥5 ✔

x=3: y=54/3=18 ≥5 ✔

x=6: y=54/6=9 ≥5 ✔

x=9: y=54/9=6 ≥5 ✔

x=18: y=54/18=3, which is less than 5. So this is invalid.

Similarly, x=27: y=2, invalid. x=54: y=1, invalid. So indeed, only x=2,3,6,9 satisfy both x≥2 and y≥5.

Wait, but hold on. Let me check if there are other divisors of 54 between 2 and 10 that I might have missed. The divisors of 54 are 1,2,3,6,9,18,27,54. So between 2 and 10, we have 2,3,6,9. So that's four values. Therefore, the number of possible x is 4?

But wait, let me think again. Is there a possibility that x could be something else? For example, maybe if x is not a divisor of 54? Wait, but if x is not a divisor of 54, then y would not be an integer. But the problem says "I deal all of the cards to x players, with each player getting y cards." So, that implies that y must be an integer. Therefore, x must divide 54. Therefore, only divisors of 54 are possible. Therefore, the possible x's are only those divisors of 54 that satisfy x ≥2 and y=54/x ≥5. Which gives us x=2,3,6,9. Therefore, 4 possible values. Therefore, the answer is 4?

Wait, but let me check again. Let me think if there's another way to approach this. Maybe I made a mistake in the inequality?

Original equation: x*y=54, with x ≥2, y ≥5. So, x must be a divisor of 54, and also y=54/x ≥5. So, 54/x ≥5 => x ≤54/5=10.8. Since x is integer, x ≤10. So x must be a divisor of 54 between 2 and 10 inclusive. Divisors of 54 in that range are 2,3,6,9. So that's four values. Therefore, answer is 4? But let me check again if there are more divisors. For example, 54 divided by 2 is 27, which is okay. Divided by 3 is 18, okay. Divided by 6 is 9, okay. Divided by 9 is 6, okay. Divided by 10? Wait, 10 is not a divisor of 54. 54 divided by 10 is 5.4, which is not integer. So, no. So, yes, only 2,3,6,9. Therefore, 4 possible values. Therefore, the answer is 4?

But wait, hold on. Let me check if there's another way. Maybe considering that y must be at least 5, so for each x, y=54/x must be integer and ≥5. So, equivalently, x must be a divisor of 54, and x ≤54/5=10.8. So x can be 2,3,6,9. So 4 values. Therefore, answer is 4. But let me check again if there are other divisors between 2 and 10. Let me list all divisors again:

Divisors of 54: 1, 2, 3, 6, 9, 18, 27, 54. So between 2 and 10, inclusive, we have 2,3,6,9. So that's four numbers. Therefore, answer is 4. Therefore, the number of possible values of x is 4. Therefore, the answer is \boxed{4}?

Wait, but wait. Let me check if there's a mistake here. For example, if x=4, then y=54/4=13.5, which is not integer. Similarly, x=5: 54/5=10.8, not integer. x=7: 54/7≈7.71, not integer. x=8: 54/8=6.75, not integer. x=10: 54/10=5.4, not integer. So none of these are integers. Therefore, only the divisors. Therefore, yes, only 2,3,6,9. Therefore, 4 values. Therefore, answer is 4. But let me check again if the problem says "each player getting y cards", so y must be integer. Therefore, x must divide 54. Therefore, yes. Therefore, answer is 4. But wait, let me check once more.

Alternatively, maybe I need to consider that y must be at least 5, so for each x, y=54/x ≥5. Therefore, x ≤54/5=10.8. So x can be up to 10. So x can be 2,3,6,9. So 4 values. Therefore, answer is 4. Therefore, I think that is correct.

But let me check if there's a different interpretation. For example, maybe the problem allows for some cards to be left over? But the problem says "I deal all of the cards to x players", so all cards are dealt, so y must be integer. Therefore, x must divide 54. Therefore, the answer is 4. Therefore, I think the answer is 4. But let me check once again.

Wait, but maybe I made a mistake in the divisors. Let me list all divisors again. 54 divided by 1 is 54. Divided by 2 is 27. Divided by 3 is 18. Divided by 6 is 9. Divided by 9 is 6. Divided by 18 is 3. Divided by 27 is 2. Divided by 54 is 1. So divisors are 1,2,3,6,9,18,27,54. So yes, that's correct. So between 2 and 10, the divisors are 2,3,6,9. So four numbers. Therefore, answer is 4. Therefore, I think that is correct.

But wait, hold on. Let me check if x=1 is allowed? But the problem says x is at least 2, so x=1 is excluded. Therefore, yes. Therefore, answer is 4. Therefore, I think the answer is 4. Therefore, the number of possible values of x is 4. Therefore, the answer is \boxed{4}.

Wait, but hold on. Let me check once more. Maybe I need to check if y is at least 5. For x=2, y=27, which is okay. For x=3, y=18, okay. For x=6, y=9, okay. For x=9, y=6, okay. So all of these satisfy y ≥5. If x=1, y=54, but x must be at least 2. If x=18, y=3, which is less than 5. So invalid. Similarly for higher x. Therefore, yes, only 4 values. Therefore, answer is 4. Therefore, I think that is correct.

But wait, hold on. Let me check if there's another way to think about this. For example, if we think of possible values of y. Since y must be at least 5, and x must be at least 2, then y can be from 5 up to 54/2=27. But since x must be integer, y must divide 54. Wait, no. Wait, if x=54/y, so y must divide 54. Therefore, y must be a divisor of 54, and y ≥5. So, let me think. If I approach it from the y side, then y must be a divisor of 54, y ≥5, and x=54/y must be integer and ≥2. Therefore, same thing. So, divisors of 54 that are ≥5. Let me list them:

Divisors of 54: 1,2,3,6,9,18,27,54. So divisors ≥5 are 6,9,18,27,54. Then, for each of these y's, check if x=54/y is ≥2.

For y=6: x=54/6=9 ≥2 ✔

y=9: x=54/9=6 ≥2 ✔

y=18: x=54/18=3 ≥2 ✔

y=27: x=54/27=2 ≥2 ✔

y=54: x=54/54=1, which is less than 2. So invalid.

Therefore, the valid y's are 6,9,18,27. Therefore, corresponding x's are 9,6,3,2. Which is the same as before. Therefore, the number of x's is 4. Therefore, same answer. Therefore, answer is 4. Therefore, I think that is correct.

Therefore, I think the answer is 4. Therefore, the number of possible values of x is 4. Therefore, the answer is \boxed{4}.

**Final Answer**
\boxed{4}

To determine how many possible values of $ x $ (the number of players) are valid under the given constraints, we begin by analyzing the relationship between the number of players and the number of cards each player receives.

---

### Step 1: Understand the Relationship

We are told that:

- There are **54 cards** in the deck.
- All cards are dealt to $ x $ players.
- Each player receives $ y $ cards.
- Therefore, the total number of cards is:

$$
x \cdot y = 54
$$

This implies that $ x $ must be a **divisor of 54**, and $ y = \frac{54}{x} $ must be an integer.

---

### Step 2: Apply the Constraints

We are given two constraints:

- $ x \geq 2 $
- $ y \geq 5 $

From the equation $ y = \frac{54}{x} $, the second constraint becomes:

$$
\frac{54}{x} \geq 5 \Rightarrow x \leq \frac{54}{5} = 10.8
$$

Since $ x $ must be an integer, we conclude:

$$
2 \leq x \leq 10
$$

Now, we find all **divisors of 54** that lie in this range.

---

### Step 3: List the Divisors of 54

The prime factorization of 54 is:

$$
54 = 2 \cdot 3^3
$$

From this, we can list all the divisors of 54:

$$
1, 2, 3, 6, 9, 18, 27, 54
$$

Now, we filter these to find the ones that satisfy:

- $ x \geq 2 $
- $ x \leq 10 $

This gives us:

$$
x = 2, 3, 6, 9
$$

---

### Step 4: Verify the Corresponding $ y $ Values

For each of these $ x $ values, we compute $ y = \frac{54}{x} $:

- $ x = 2 \Rightarrow y = 27 $
- $ x = 3 \Rightarrow y = 18 $
- $ x = 6 \Rightarrow y = 9 $
- $ x = 9 \Rightarrow y = 6 $

All of these values of $ y $ are **greater than or equal to 5**, satisfying the constraint.

---

### Step 5: Final Count

We have found **4 valid values** of $ x $ that satisfy all the conditions:

$$
x = 2, 3, 6, 9
$$

---

### Final Answer

$$
\boxed{4}
$$

Solution: We want $xy=54=2 \cdot 3^3$ such that $x$ is at least $2$ and $y$ is at least  $5$. Thus, the possible combinations $(x,y)$ are $(2,27)$, $(3,18)$, $(6,9)$, and $(9,6)$. There are $\boxed{4}$ such combinations.