Explanation:
Okay, so I need to find the minimum value of the expression $\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$ given that $a + b + c + d = 1$ where all variables are positive real numbers. Alright, let's start by understanding the problem.

First, there are four variables with a constraint that their sum is 1. The function we need to minimize is a sum of reciprocals with different coefficients. Since the coefficients are different, it's not symmetric, so maybe I can't use the same method as the AM-GM inequality directly? Or maybe I can?

Let me recall the AM-GM inequality. The Arithmetic Mean - Geometric Mean inequality states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. Equality holds when all numbers are equal. So, maybe I can apply AM-GM here somehow. However, the problem is that the coefficients in the expression are different. So, perhaps I need to adjust the variables or use weighted AM-GM?

Alternatively, maybe Lagrange multipliers? Since this is a constrained optimization problem. Let me think about both approaches.

Starting with Lagrange multipliers. Let me set up the function to minimize and the constraint.

Let $f(a, b, c, d) = \frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$, and the constraint is $g(a, b, c, d) = a + b + c + d - 1 = 0$.

Then, the method of Lagrange multipliers says that at the minimum, the gradient of f is proportional to the gradient of g. That is,

$$
\nabla f = \lambda \nabla g
$$

Calculating the partial derivatives.

For variable a:

$$
\frac{\partial f}{\partial a} = -\frac{1}{a^2}
$$
$$
\frac{\partial g}{\partial a} = 1
$$

Similarly for variable b:

$$
\frac{\partial f}{\partial b} = -\frac{1}{b^2}
$$
$$
\frac{\partial g}{\partial b} = 1
$$

For variable c:

$$
\frac{\partial f}{\partial c} = -\frac{4}{c^2}
$$
$$
\frac{\partial g}{\partial c} = 1
$$

For variable d:

$$
\frac{\partial f}{\partial d} = -\frac{16}{d^2}
$$
$$
\frac{\partial g}{\partial d} = 1
$$

Therefore, setting up the equations:

$$
-\frac{1}{a^2} = \lambda \cdot 1 \quad (1)
$$
$$
-\frac{1}{b^2} = \lambda \cdot 1 \quad (2)
$$
$$
-\frac{4}{c^2} = \lambda \cdot 1 \quad (3)
$$
$$
-\frac{16}{d^2} = \lambda \cdot 1 \quad (4)
$$

So from equations (1) and (2):

$$
-\frac{1}{a^2} = -\frac{1}{b^2} \implies \frac{1}{a^2} = \frac{1}{b^2} \implies a = b
$$

Since a and b are positive, we don't have to worry about negative roots.

Similarly, from equations (1) and (3):

$$
-\frac{1}{a^2} = -\frac{4}{c^2} \implies \frac{1}{a^2} = \frac{4}{c^2} \implies c^2 = 4a^2 \implies c = 2a
$$

Similarly, from equations (1) and (4):

$$
-\frac{1}{a^2} = -\frac{16}{d^2} \implies \frac{1}{a^2} = \frac{16}{d^2} \implies d^2 = 16a^2 \implies d = 4a
$$

So, from these relations, we can express b, c, d in terms of a. Since a = b, c = 2a, d = 4a. Therefore, let's substitute these into the constraint equation.

Original constraint:

$$
a + b + c + d = 1
$$

Substituting:

$$
a + a + 2a + 4a = 1 \implies 8a = 1 \implies a = \frac{1}{8}
$$

Therefore, the variables are:

- $a = \frac{1}{8}$
- $b = \frac{1}{8}$
- $c = 2 \times \frac{1}{8} = \frac{1}{4}$
- $d = 4 \times \frac{1}{8} = \frac{1}{2}$

Now, compute the value of the expression:

$$
\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} = \frac{1}{1/8} + \frac{1}{1/8} + \frac{4}{1/4} + \frac{16}{1/2}
$$

Calculating each term:

- $\frac{1}{1/8} = 8$
- $\frac{1}{1/8} = 8$
- $\frac{4}{1/4} = 16$
- $\frac{16}{1/2} = 32$

Adding them up: 8 + 8 + 16 + 32 = 64. So the minimum value is 64? Wait, let me check my calculations again.

Wait, 8 + 8 is 16, 16 + 16 is 32, and 32 + 32 is 64. Yes, that's correct. Hmm, okay. But before accepting that, let me verify if this is indeed the minimum. Maybe there's a mistake in assuming the Lagrange multipliers?

Alternatively, maybe there's another way to approach this problem, such as using the Cauchy-Schwarz inequality or AM-GM.

Let me try the AM-GM approach. Let's think about how to apply AM-GM here. Since we have different coefficients, maybe we can split the terms in such a way that we can apply AM-GM.

Alternatively, think about using weighted AM-GM. Let me recall that the weighted AM-GM inequality states that for positive real numbers and positive weights summing to 1, the weighted arithmetic mean is greater than or equal to the weighted geometric mean.

But perhaps another approach is to consider the terms in the expression and the constraint. Let me consider variables with different coefficients. Let me see.

Alternatively, maybe we can make substitutions. Let me think. Let me try to consider variables that can be combined.

Let me note that in the Lagrange method, we found that a = b, c = 2a, d = 4a. So, there's a pattern here. Maybe if I think of the terms in the expression as being related to the coefficients in the denominators. Let me see:

Original expression: 1/a + 1/b + 4/c + 16/d.

Notice that 1, 1, 4, 16 are squares of 1, 1, 2, 4. Interesting. Also, in the constraint, the variables are a, b, c, d. If I think of the coefficients in the expression as squares, maybe there's a way to pair them with the variables?

Alternatively, maybe using the Cauchy-Schwarz inequality. Let me think.

Let me recall that for vectors u and v, (u ⋅ v)^2 ≤ (u ⋅ u)(v ⋅ v). But not sure how to apply here.

Alternatively, think of the expression as a sum of terms, and relate them to the constraint.

Alternatively, use the method of substitution. Since we have a + b + c + d = 1, maybe express one variable in terms of others. But with four variables, that might get complicated. However, since in the Lagrange method we found relations between variables, maybe that gives us the optimal.

But let me try to check if this is indeed the minimum. Let me test with different values.

Suppose I take a = b = 1/8, c = 1/4, d = 1/2. Then the total sum is 1/8 + 1/8 + 1/4 + 1/2 = 1/8 + 1/8 + 2/8 + 4/8 = (1 + 1 + 2 + 4)/8 = 8/8 = 1. That's correct.

Now, let me try to perturb one variable. For example, suppose I take a = 1/8 + ε, b = 1/8 - ε, keeping a + b the same. Then, let me see how the expression changes. Wait, but I need to adjust other variables as well since the total sum is fixed. Maybe this is getting too complicated.

Alternatively, take another set of values. Suppose I set a = b = 1/6, c = 1/6, d = 1/6. Then the sum is 1/6 * 4 = 2/3, which is less than 1. So to adjust, let me make them larger. Wait, maybe not. Let's try to think of another example.

Alternatively, take a = 0.2, b = 0.2, c = 0.2, d = 0.4. Then sum is 0.2 + 0.2 + 0.2 + 0.4 = 1. Then compute the expression:

1/0.2 + 1/0.2 + 4/0.2 + 16/0.4 = 5 + 5 + 20 + 40 = 70. Which is bigger than 64. So worse.

Another example, take a = 1/8, b = 1/8, c = 1/4, d = 1/2. Which gives 64. Let me try another set. Suppose I take a = 0.1, b = 0.1, c = 0.3, d = 0.5. Then check sum: 0.1 + 0.1 + 0.3 + 0.5 = 1. Compute the expression: 1/0.1 + 1/0.1 + 4/0.3 + 16/0.5 = 10 + 10 + 13.333... + 32 ≈ 65.333... which is still higher than 64. So seems like 64 is better.

Another test case: Let me try a = 1/9, b = 1/9, c = 1/3, d = 1 - 1/9 -1/9 -1/3 = 1 - (2/9 + 3/9) = 1 - 5/9 = 4/9. Then compute:

1/(1/9) + 1/(1/9) + 4/(1/3) + 16/(4/9) = 9 + 9 + 12 + 36 = 66. Also higher.

Alternatively, take a = 1/10, b = 1/10, c = 1/2, d = 1 - 1/10 -1/10 -1/2 = 1 - (2/10 + 5/10) = 1 - 7/10 = 3/10.

Compute expression: 10 + 10 + 4/(1/2) +16/(3/10) = 20 + 8 + (160/3) ≈ 28 + 53.333... ≈ 81.333. Worse.

Alternatively, try to make c and d smaller? Wait, but since they are in denominators, making them smaller would make the terms larger. So probably not helpful. Let me try with larger c and d? For example, take a = 1/8, b = 1/8, c = 1/4 + ε, d = 1/2 - ε. Let's choose ε = 0.01, then c = 0.25 + 0.01 = 0.26, d = 0.5 - 0.01 = 0.49. Then compute expression:

1/a + 1/b + 4/c + 16/d ≈ 8 + 8 + 4/0.26 + 16/0.49 ≈ 16 + 15.3846 + 32.653 ≈ 16 + 15.38 + 32.65 ≈ 64.03. So, approximately 64.03, which is slightly higher than 64. So, that seems to confirm that the minimal value is indeed 64.

Alternatively, maybe take ε negative? If I take ε = -0.01, then c = 0.24, d = 0.51. Then compute:

4/0.24 ≈ 16.666..., and 16/0.51 ≈ 31.3725. Then total expression ≈ 16 + 16.666 + 31.372 ≈ 64.038, which is still slightly higher. So, this suggests that the minimal value is indeed achieved at a = b = 1/8, c = 1/4, d = 1/2.

Alternatively, perhaps trying another approach. Let me think again about the Lagrange multipliers. So according to that method, the critical point is at those values, and since the function tends to infinity as any variable approaches 0, this critical point is likely to be the global minimum.

Alternatively, let me try to apply the Cauchy-Schwarz inequality. Let me think of vectors. The Cauchy-Schwarz inequality states that (sum u_i v_i)^2 ≤ (sum u_i^2)(sum v_i^2). But how to apply here?

Alternatively, think of the expression $\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$ as a sum of terms. Maybe I can use the Cauchy-Schwarz in the form:

$(\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d})(a + b + c + d) \geq (1 + 1 + 2 + 4)^2$.

Wait, is this valid? Let me recall that Cauchy-Schwarz can be applied as follows:

For sequences (x1, x2, x3, x4) and (y1, y2, y3, y4), we have (x1y1 + x2y2 + x3y3 + x4y4)^2 ≤ (x1^2 + x2^2 + x3^2 + x4^2)(y1^2 + y2^2 + y3^2 + y4^2). But if we want to use it to relate sums with reciprocals, maybe we need to set up something else.

Alternatively, using the Titu's lemma, which is a form of Cauchy-Schwarz. Titu's lemma states that for positive reals, $\sum \frac{p_i^2}{q_i} \ge \frac{(\sum p_i)^2}{\sum q_i}$. Let me try that.

But our expression is $\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$. Let me write this as $\frac{1^2}{a} + \frac{1^2}{b} + \frac{2^2}{c} + \frac{4^2}{d}$. Then according to Titu's lemma:

$$
\frac{1^2}{a} + \frac{1^2}{b} + \frac{2^2}{c} + \frac{4^2}{d} \ge \frac{(1 + 1 + 2 + 4)^2}{a + b + c + d} = \frac{8^2}{1} = 64
$$

So that gives us the lower bound of 64. And since we have equality condition in Titu's lemma when $\frac{p_i}{q_i}$ is constant for all i. Wait, Titu's lemma is derived from Cauchy-Schwarz, and equality holds when $\frac{p_i}{q_i}$ is constant for all i. So in our case, for Titu's lemma, the equality condition would be $\frac{1}{a} = \frac{1}{b} = \frac{2}{c} = \frac{4}{d}$.

Let me check. Let me see:

If equality holds in Titu's lemma, then $\frac{p_1}{q_1} = \frac{p_2}{q_2} = \cdots = \frac{p_n}{q_n}$. In our case, the p_i are 1,1,2,4, and the q_i are a,b,c,d. Therefore:

$\frac{1}{a} = \frac{1}{b} = \frac{2}{c} = \frac{4}{d}$. Let me denote this common ratio as k. Therefore:

From $\frac{1}{a} = k$ implies $a = 1/k$.

Similarly, $b = 1/k$, $c = 2/k$, $d = 4/k$.

Therefore, the sum $a + b + c + d = 1/k + 1/k + 2/k + 4/k = 8/k = 1$, which implies that $k = 8$.

Therefore, a = 1/8, b = 1/8, c = 2/8 = 1/4, d = 4/8 = 1/2, which matches exactly the result from the Lagrange multipliers method!

Therefore, the minimal value is indeed 64, achieved when a = b = 1/8, c = 1/4, d = 1/2. So this confirms the earlier result. Therefore, the answer is 64.

But let me check again if Titu's lemma was applied correctly. The original expression is $\frac{1^2}{a} + \frac{1^2}{b} + \frac{2^2}{c} + \frac{4^2}{d}$, which is exactly the form required for Titu's lemma. Then, applying Titu's lemma gives the lower bound of $\frac{(1 + 1 + 2 + 4)^2}{a + b + c + d} = 64 / 1 = 64$. Therefore, the minimum value is indeed 64. This seems solid.

Alternatively, another way to think about this is using the AM-HM inequality, but since the coefficients are different, Titu's lemma is more appropriate here.

Therefore, both methods lead to the same conclusion. Therefore, I can be confident that the minimal value is 64.

Just to recap, using Titu's lemma (a specific case of Cauchy-Schwarz) gives us the lower bound directly, and the equality condition is achievable with the given constraint, hence 64 is indeed the minimal value.

So, the answer is $\boxed{64}$.

**Final Answer**
\boxed{64}

To find the **minimum value** of the expression
$$
\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}
$$
under the constraint
$$
a + b + c + d = 1,
$$
with all variables being **positive real numbers**, we can use a powerful inequality known as **Titu's Lemma** (which is a consequence of the Cauchy-Schwarz inequality).

---

### Step 1: Apply Titu's Lemma

We observe that the given expression can be rewritten as:

$$
\frac{1^2}{a} + \frac{1^2}{b} + \frac{2^2}{c} + \frac{4^2}{d}.
$$

This has the form
$$
\sum_{i=1}^{n} \frac{x_i^2}{y_i},
$$
where $ x_i = 1, 1, 2, 4 $ and $ y_i = a, b, c, d $. According to **Titu's Lemma**, we have:

$$
\sum_{i=1}^n \frac{x_i^2}{y_i} \geq \frac{(x_1 + x_2 + \cdots + x_n)^2}{y_1 + y_2 + \cdots + y_n}.
$$

Applying this inequality:

$$
\frac{1^2}{a} + \frac{1^2}{b} + \frac{2^2}{c} + \frac{4^2}{d} \geq \frac{(1 + 1 + 2 + 4)^2}{a + b + c + d}.
$$

Since $ a + b + c + d = 1 $, the denominator is 1, so the inequality becomes:

$$
\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} \geq \frac{(8)^2}{1} = 64.
$$

---

### Step 2: Achievability of the Bound

To check whether equality is achievable, we recall that **equality in Titu's Lemma** holds if and only if the ratio between the numerators and denominators is constant. That is:

$$
\frac{1}{a} = \frac{1}{b} = \frac{2}{c} = \frac{4}{d}.
$$

Let us denote this common ratio as $ k $. Then:

$$
a = \frac{1}{k}, \quad b = \frac{1}{k}, \quad c = \frac{2}{k}, \quad d = \frac{4}{k}.
$$

Substituting into the constraint:

$$
a + b + c + d = \frac{1}{k} + \frac{1}{k} + \frac{2}{k} + \frac{4}{k} = \frac{8}{k} = 1 \Rightarrow k = 8.
$$

So we have:

$$
a = b = \frac{1}{8}, \quad c = \frac{1}{4}, \quad d = \frac{1}{2}.
$$

Substituting into the expression:

$$
\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} = 8 + 8 + 16 + 32 = 64.
$$

---

### Final Answer

Thus, the **minimum value** of the expression is:

$$
\boxed{64}
$$

[Assistant]
Explanation:
Okay, so I need to find the minimum value of the expression $\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$ given that $a + b + c + d = 1$ where all variables are positive real numbers. Alright, let's start by understanding the problem.

First, there are four variables with a constraint that their sum is 1. The function we need to minimize is a sum of reciprocals with different coefficients. Since the coefficients are different, it's not symmetric, so maybe I can't use the same method as the AM-GM inequality directly? Or maybe I can?

Let me recall the AM-GM inequality. The Arithmetic Mean - Geometric Mean inequality states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. Equality holds when all numbers are equal. So, maybe I can apply AM-GM here somehow. However, the problem is that the coefficients in the expression are different. So, perhaps I need to adjust the variables or use weighted AM-GM?

Alternatively, maybe Lagrange multipliers? Since this is a constrained optimization problem. Let me think about both approaches.

Starting with Lagrange multipliers. Let me set up the function to minimize and the constraint.

Let $f(a, b, c, d) = \frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$, and the constraint is $g(a, b, c, d) = a + b + c + d - 1 = 0$.

Then, the method of Lagrange multipliers says that at the minimum, the gradient of f is proportional to the gradient of g. That is,

$$
\nabla f = \lambda \nabla g
$$

Calculating the partial derivatives.

For variable a:

$$
\frac{\partial f}{\partial a} = -\frac{1}{a^2}
$$
$$
\frac{\partial g}{\partial a} = 1
$$

Similarly for variable b:

$$
\frac{\partial f}{\partial b} = -\frac{1}{b^2}
$$
$$
\frac{\partial g}{\partial b} = 1
$$

For variable c:

$$
\frac{\partial f}{\partial c} = -\frac{4}{c^2}
$$
$$
\frac{\partial g}{\partial c} = 1
$$

For variable d:

$$
\frac{\partial f}{\partial d} = -\frac{16}{d^2}
$$
$$
\frac{\partial g}{\partial d} = 1
$$

Therefore, setting up the equations:

$$
-\frac{1}{a^2} = \lambda \cdot 1 \quad (1)
$$
$$
-\frac{1}{b^2} = \lambda \cdot 1 \quad (2)
$$
$$
-\frac{4}{c^2} = \lambda \cdot 1 \quad (3)
$$
$$
-\frac{16}{d^2} = \lambda \cdot 1 \quad (4)
$$

So from equations (1) and (2):

$$
-\frac{1}{a^2} = -\frac{1}{b^2} \implies \frac{1}{a^2} = \frac{1}{b^2} \implies a = b
$$

Since a and b are positive, we don't have to worry about negative roots.

Similarly, from equations (1) and (3):

$$
-\frac{1}{a^2} = -\frac{4}{c^2} \implies \frac{1}{a^2} = \frac{4}{c^2} \implies c^2 = 4a^2 \implies c = 2a
$$

Similarly, from equations (1) and (4):

$$
-\frac{1}{a^2} = -\frac{16}{d^2} \implies \frac{1}{a^2} = \frac{16}{d^2} \implies d^2 = 16a^2 \implies d = 4a
$$

So, from these relations, we can express b, c, d in terms of a. Since a = b, c = 2a, d = 4a. Therefore, let's substitute these into the constraint equation.

Original constraint:

$$
a + b + c + d = 1
$$

Substituting:

$$
a + a + 2a + 4a = 1 \implies 8a = 1 \implies a = \frac{1}{8}
$$

Therefore, the variables are:

- $a = \frac{1}{8}$
- $b = \frac{1}{8}$
- $c = 2 \times \frac{1}{8} = \frac{1}{4}$
- $d = 4 \times \frac{1}{8} = \frac{1}{2}$

Now, compute the value of the expression:

$$
\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} = \frac{1}{1/8} + \frac{1}{1/8} + \frac{4}{1/4} + \frac{16}{1/2}
$$

Calculating each term:

- $\frac{1}{1/8} = 8$
- $\frac{1}{1/8} = 8$
- $\frac{4}{1/4} = 16$
- $\frac{16}{1/2} = 32$

Adding them up: 8 + 8 + 16 + 32 = 64. So the minimum value is 64? Wait, let me check my calculations again.

Wait, 8 + 8 is 16, 16 + 16 is 32, and 32 + 32 is 64. Yes, that's correct. Hmm, okay. But before accepting that, let me verify if this is indeed the minimum. Maybe there's a mistake in assuming the Lagrange multipliers?

Alternatively, maybe there's another way to approach this problem, such as using the Cauchy-Schwarz inequality or AM-GM.

Let me try the AM-GM approach. Let's think about how to apply AM-GM here. Since we have different coefficients, maybe we can split the terms in such a way that we can apply AM-GM.

Alternatively, think about using weighted AM-GM. Let me recall that the weighted AM-GM inequality states that for positive real numbers and positive weights summing to 1, the weighted arithmetic mean is greater than or equal to the weighted geometric mean.

But perhaps another approach is to consider the terms in the expression and the constraint. Let me consider variables with different coefficients. Let me see.

Alternatively, maybe we can make substitutions. Let me think. Let me try to consider variables that can be combined.

Let me note that in the Lagrange method, we found that a = b, c = 2a, d = 4a. So, there's a pattern here. Maybe if I think of the terms in the expression as being related to the coefficients in the denominators. Let me see:

Original expression: 1/a + 1/b + 4/c + 16/d.

Notice that 1, 1, 4, 16 are squares of 1, 1, 2, 4. Interesting. Also, in the constraint, the variables are a, b, c, d. If I think of the coefficients in the expression as squares, maybe there's a way to pair them with the variables?

Alternatively, maybe using the Cauchy-Schwarz inequality. Let me think.

Let me recall that for vectors u and v, (u ⋅ v)^2 ≤ (u ⋅ u)(v ⋅ v). But not sure how to apply here.

Alternatively, think of the expression as a sum of terms, and relate them to the constraint.

Alternatively, use the method of substitution. Since we have a + b + c + d = 1, maybe express one variable in terms of others. But with four variables, that might get complicated. However, since in the Lagrange method we found relations between variables, maybe that gives us the optimal.

But let me try to check if this is indeed the minimum. Let me test with different values.

Suppose I take a = b = 1/8, c = 1/4, d = 1/2. Then the total sum is 1/8 + 1/8 + 1/4 + 1/2 = 1/8 + 1/8 + 2/8 + 4/8 = (1 + 1 + 2 + 4)/8 = 8/8 = 1. That's correct.

Now, let me try to perturb one variable. For example, suppose I take a = 1/8 + ε, b = 1/8 - ε, keeping a + b the same. Then, let me see how the expression changes. Wait, but I need to adjust other variables as well since the total sum is fixed. Maybe this is getting too complicated.

Alternatively, take another set of values. Suppose I set a = b = 1/6, c = 1/6, d = 1/6. Then the sum is 1/6 * 4 = 2/3, which is less than 1. So to adjust, let me make them larger. Wait, maybe not. Let's try to think of another example.

Alternatively, take a = 0.2, b = 0.2, c = 0.2, d = 0.4. Then sum is 0.2 + 0.2 + 0.2 + 0.4 = 1. Then compute the expression:

1/0.2 + 1/0.2 + 4/0.2 + 16/0.4 = 5 + 5 + 20 + 40 = 70. Which is bigger than 64. So worse.

Another example, take a = 1/8, b = 1/8, c = 1/4, d = 1/2. Which gives 64. Let me try another set. Suppose I take a = 0.1, b = 0.1, c = 0.3, d = 0.5. Then check sum: 0.1 + 0.1 + 0.3 + 0.5 = 1. Compute the expression: 1/0.1 + 1/0.1 + 4/0.3 + 16/0.5 = 10 + 10 + 13.333... + 32 ≈ 65.333... which is still higher than 64. So seems like 64 is better.

Another test case: Let me try a = 1/9, b = 1/9, c = 1/3, d = 1 - 1/9 -1/9 -1/3 = 1 - (2/9 + 3/9) = 1 - 5/9 = 4/9. Then compute:

1/(1/9) + 1/(1/9) + 4/(1/3) + 16/(4/9) = 9 + 9 + 12 + 36 = 66. Also higher.

Alternatively, take a = 1/10, b = 1/10, c = 1/2, d = 1 - 1/10 -1/10 -1/2 = 1 - (2/10 + 5/10) = 1 - 7/10 = 3/10.

Compute expression: 10 + 10 + 4/(1/2) +16/(3/10) = 20 + 8 + (160/3) ≈ 28 + 53.333... ≈ 81.333. Worse.

Alternatively, try to make c and d smaller? Wait, but since they are in denominators, making them smaller would make the terms larger. So probably not helpful. Let me try with larger c and d? For example, take a = 1/8, b = 1/8, c = 1/4 + ε, d = 1/2 - ε. Let's choose ε = 0.01, then c = 0.25 + 0.01 = 0.26, d = 0.5 - 0.01 = 0.49. Then compute expression:

1/a + 1/b + 4/c + 16/d ≈ 8 + 8 + 4/0.26 + 16/0.49 ≈ 16 + 15.3846 + 32.653 ≈ 16 + 15.38 + 32.65 ≈ 64.03. So, approximately 64.03, which is slightly higher than 64. So, that seems to confirm that the minimal value is indeed 64.

Alternatively, maybe take ε negative? If I take ε = -0.01, then c = 0.24, d = 0.51. Then compute:

4/0.24 ≈ 16.666..., and 16/0.51 ≈ 31.3725. Then total expression ≈ 16 + 16.666 + 31.372 ≈ 64.038, which is still slightly higher. So, this suggests that the minimal value is indeed achieved at a = b = 1/8, c = 1/4, d = 1/2.

Alternatively, perhaps trying another approach. Let me think again about the Lagrange multipliers. So according to that method, the critical point is at those values, and since the function tends to infinity as any variable approaches 0, this critical point is likely to be the global minimum.

Alternatively, let me try to apply the Cauchy-Schwarz inequality. Let me think of vectors. The Cauchy-Schwarz inequality states that (sum u_i v_i)^2 ≤ (sum u_i^2)(sum v_i^2). But how to apply here?

Alternatively, think of the expression $\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$ as a sum of terms. Maybe I can use the Cauchy-Schwarz in the form:

$(\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d})(a + b + c + d) \geq (1 + 1 + 2 + 4)^2$.

Wait, is this valid? Let me recall that Cauchy-Schwarz can be applied as follows:

For sequences (x1, x2, x3, x4) and (y1, y2, y3, y4), we have (x1y1 + x2y2 + x3y3 + x4y4)^2 ≤ (x1^2 + x2^2 + x3^2 + x4^2)(y1^2 + y2^2 + y3^2 + y4^2). But if we want to use it to relate sums with reciprocals, maybe we need to set up something else.

Alternatively, using the Titu's lemma, which is a form of Cauchy-Schwarz. Titu's lemma states that for positive reals, $\sum \frac{p_i^2}{q_i} \ge \frac{(\sum p_i)^2}{\sum q_i}$. Let me try that.

But our expression is $\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$. Let me write this as $\frac{1^2}{a} + \frac{1^2}{b} + \frac{2^2}{c} + \frac{4^2}{d}$. Then according to Titu's lemma:

$$
\frac{1^2}{a} + \frac{1^2}{b} + \frac{2^2}{c} + \frac{4^2}{d} \ge \frac{(1 + 1 + 2 + 4)^2}{a + b + c + d} = \frac{8^2}{1} = 64
$$

So that gives us the lower bound of 64. And since we have equality condition in Titu's lemma when $\frac{p_i}{q_i}$ is constant for all i. Wait, Titu's lemma is derived from Cauchy-Schwarz, and equality holds when $\frac{p_i}{q_i}$ is constant for all i. So in our case, for Titu's lemma, the equality condition would be $\frac{1}{a} = \frac{1}{b} = \frac{2}{c} = \frac{4}{d}$.

Let me check. Let me see:

If equality holds in Titu's lemma, then $\frac{p_1}{q_1} = \frac{p_2}{q_2} = \cdots = \frac{p_n}{q_n}$. In our case, the p_i are 1,1,2,4, and the q_i are a,b,c,d. Therefore:

$\frac{1}{a} = \frac{1}{b} = \frac{2}{c} = \frac{4}{d}$. Let me denote this common ratio as k. Therefore:

From $\frac{1}{a} = k$ implies $a = 1/k$.

Similarly, $b = 1/k$, $c = 2/k$, $d = 4/k$.

Therefore, the sum $a + b + c + d = 1/k + 1/k + 2/k + 4/k = 8/k = 1$, which implies that $k = 8$.

Therefore, a = 1/8, b = 1/8, c = 2/8 = 1/4, d = 4/8 = 1/2, which matches exactly the result from the Lagrange multipliers method!

Therefore, the minimal value is indeed 64, achieved when a = b = 1/8, c = 1/4, d = 1/2. So this confirms the earlier result. Therefore, the answer is 64.

But let me check again if Titu's lemma was applied correctly. The original expression is $\frac{1^2}{a} + \frac{1^2}{b} + \frac{2^2}{c} + \frac{4^2}{d}$, which is exactly the form required for Titu's lemma. Then, applying Titu's lemma gives the lower bound of $\frac{(1 + 1 + 2 + 4)^2}{a + b + c + d} = 64 / 1 = 64$. Therefore, the minimum value is indeed 64. This seems solid.

Alternatively, another way to think about this is using the AM-HM inequality, but since the coefficients are different, Titu's lemma is more appropriate here.

Therefore, both methods lead to the same conclusion. Therefore, I can be confident that the minimal value is 64.

Just to recap, using Titu's lemma (a specific case of Cauchy-Schwarz) gives us the lower bound directly, and the equality condition is achievable with the given constraint, hence 64 is indeed the minimal value.

So, the answer is $\boxed{64}$.

**Final Answer**
\boxed{64}

To find the **minimum value** of the expression
$$
\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}
$$
under the constraint
$$
a + b + c + d = 1,
$$
with all variables being **positive real numbers**, we can use a powerful inequality known as **Titu's Lemma** (which is a consequence of the Cauchy-Schwarz inequality).

---

### Step 1: Apply Titu's Lemma

We observe that the given expression can be rewritten as:

$$
\frac{1^2}{a} + \frac{1^2}{b} + \frac{2^2}{c} + \frac{4^2}{d}.
$$

This has the form
$$
\sum_{i=1}^{n} \frac{x_i^2}{y_i},
$$
where $ x_i = 1, 1, 2, 4 $ and $ y_i = a, b, c, d $. According to **Titu's Lemma**, we have:

$$
\sum_{i=1}^n \frac{x_i^2}{y_i} \geq \frac{(x_1 + x_2 + \cdots + x_n)^2}{y_1 + y_2 + \cdots + y_n}.
$$

Applying this inequality:

$$
\frac{1^2}{a} + \frac{1^2}{b} + \frac{2^2}{c} + \frac{4^2}{d} \geq \frac{(1 + 1 + 2 + 4)^2}{a + b + c + d}.
$$

Since $ a + b + c + d = 1 $, the denominator is 1, so the inequality becomes:

$$
\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} \geq \frac{(8)^2}{1} = 64.
$$

---

### Step 2: Achievability of the Bound

To check whether equality is achievable, we recall that **equality in Titu's Lemma** holds if and only if the ratio between the numerators and denominators is constant. That is:

$$
\frac{1}{a} = \frac{1}{b} = \frac{2}{c} = \frac{4}{d}.
$$

Let us denote this common ratio as $ k $. Then:

$$
a = \frac{1}{k}, \quad b = \frac{1}{k}, \quad c = \frac{2}{k}, \quad d = \frac{4}{k}.
$$

Substituting into the constraint:

$$
a + b + c + d = \frac{1}{k} + \frac{1}{k} + \frac{2}{k} + \frac{4}{k} = \frac{8}{k} = 1 \Rightarrow k = 8.
$$

So we have:

$$
a = b = \frac{1}{8}, \quad c = \frac{1}{4}, \quad d = \frac{1}{2}.
$$

Substituting into the expression:

$$
\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} = 8 + 8 + 16 + 32 = 64.
$$

---

### Final Answer

Thus, the **minimum value** of the expression is:

$$
\boxed{64}
$$

Solution:
By Cauchy-Schwarz,
\[(a + b + c + d) \left( \frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} \right) \ge (1 + 1 + 2 + 4)^2 = 64.\]Equality occurs when $a = b = \frac{c}{2} = \frac{d}{4}$ and $a + b + c + d = 1.$  We can solve to get $a = \frac{1}{8},$ $b = \frac{1}{8},$ $c = \frac{1}{4},$ and $d = \frac{1}{2},$ so the minimum value is $\boxed{64}.$


Explanation:
Okay, so I need to find the range of the function $ f(x) = (\arccos x)^2 + (\arcsin x)^2 $. Hmm, let me start by recalling some basics about inverse trigonometric functions.

First, I know that the domains of both $ \arccos x $ and $ \arcsin x $ are $ [-1, 1] $. So, the domain of $ f(x) $ is also $ [-1, 1] $. That's straightforward. Now, I need to figure out the range of this function.

I remember that there's a relationship between $ \arccos x $ and $ \arcsin x $. Specifically, $ \arccos x + \arcsin x = \frac{\pi}{2} $ for all $ x $ in $ [-1, 1] $. Let me verify that... If I take $ \arccos x + \arcsin x $, let me denote $ \theta = \arcsin x $. Then $ \sin \theta = x $, and $ \theta $ is between $ -\frac{\pi}{2} $ and $ \frac{\pi}{2} $, but since $ \arcsin x $ is defined to return values in $ [-\frac{\pi}{2}, \frac{\pi}{2}] $. Wait, but $ \arccos x $ is in $ [0, \pi] $. So adding them together...

Alternatively, maybe I can use the identity. Let me think, if $ \theta = \arcsin x $, then $ \arccos x = \frac{\pi}{2} - \theta $. Because $ \cos\left(\frac{\pi}{2} - \theta\right) = \sin \theta = x $. Yes, so $ \arccos x = \frac{\pi}{2} - \arcsin x $. Therefore, their sum is $ \frac{\pi}{2} $.

So that means $ (\arccos x)^2 + (\arcsin x)^2 $ can be rewritten using this identity. Let me try substituting $ \arccos x = \frac{\pi}{2} - \arcsin x $ into the function.

Let me denote $ y = \arcsin x $. Then $ \arccos x = \frac{\pi}{2} - y $, so the function becomes:

$ f(x) = \left( \frac{\pi}{2} - y \right)^2 + y^2 $.

Since $ y = \arcsin x $, and $ x \in [-1, 1] $, then $ y \in [-\frac{\pi}{2}, \frac{\pi}{2}] $. However, $ \arcsin x $ actually returns values in $ [-\frac{\pi}{2}, \frac{\pi}{2}] $, but since $ \arccos x $ is in $ [0, \pi] $, but we already used that relationship. Wait, but if $ y $ is in $ [-\frac{\pi}{2}, \frac{\pi}{2}] $, then $ \frac{\pi}{2} - y $ would be in $ [\frac{\pi}{2} - \frac{\pi}{2}, \frac{\pi}{2} - (-\frac{\pi}{2})] = [0, \pi] $, which matches the range of $ \arccos x $. So that substitution is valid.

So, substituting, we have:

$ f(x) = \left( \frac{\pi}{2} - y \right)^2 + y^2 $.

Let me expand that expression:

First, square the terms:

$ \left( \frac{\pi}{2} - y \right)^2 = \left( \frac{\pi}{2} \right)^2 - \pi y + y^2 $.

Therefore, adding $ y^2 $:

$ f(x) = \left( \frac{\pi^2}{4} - \pi y + y^2 \right) + y^2 = \frac{\pi^2}{4} - \pi y + 2y^2 $.

So, simplifying, the function becomes $ f(y) = 2y^2 - \pi y + \frac{\pi^2}{4} $, where $ y \in [-\frac{\pi}{2}, \frac{\pi}{2}] $.

Now, since this is a quadratic in terms of $ y $, I can analyze its minimum and maximum values over the interval $ y \in [-\frac{\pi}{2}, \frac{\pi}{2}] $.

Let me write the quadratic again: $ f(y) = 2y^2 - \pi y + \frac{\pi^2}{4} $. To find its extrema, I can find its derivative and critical points, or use vertex formula. Since it's a quadratic, we know it's a parabola. The coefficient of $ y^2 $ is 2, which is positive, so the parabola opens upwards. Therefore, it has a minimum at its vertex and maximum at one of the endpoints of the interval.

So, first, let's find the vertex. The vertex of a quadratic $ ay^2 + by + c $ is at $ y = -\frac{b}{2a} $.

Here, $ a = 2 $, $ b = -\pi $, so the vertex is at:

$ y = -(-\pi)/(2*2) = \pi/4 $.

So the minimum occurs at $ y = \pi/4 $. Now, I need to check if this point is within the interval $ [-\frac{\pi}{2}, \frac{\pi}{2}] $. Well, $ \pi/4 $ is approximately 0.785 radians, and $ \pi/2 $ is approximately 1.5708 radians. So yes, $ \pi/4 $ is within the interval. Therefore, the minimum value of $ f(y) $ is at $ y = \pi/4 $, and the maximum occurs at one of the endpoints, either $ y = -\frac{\pi}{2} $ or $ y = \frac{\pi}{2} $.

Therefore, to find the range of $ f(x) $, I need to compute $ f(\pi/4) $, $ f(-\frac{\pi}{2}) $, and $ f(\frac{\pi}{2}) $, then determine the minimum and maximum values.

Let me compute each of these.

First, compute $ f(\pi/4) $:

$ f(\pi/4) = 2(\pi/4)^2 - \pi(\pi/4) + \frac{\pi^2}{4} $.

Let me compute each term:

First term: $ 2*(\pi^2/16) = \pi^2/8 $.

Second term: $ - \pi*(\pi/4) = - \pi^2/4 $.

Third term: $ \pi^2/4 $.

Adding them together:

$ \pi^2/8 - \pi^2/4 + \pi^2/4 = \pi^2/8 $. The $ -\pi^2/4 $ and $ +\pi^2/4 $ cancel out, leaving $ \pi^2/8 $.

So, the minimum value is $ \pi^2/8 $.

Now, compute $ f(-\frac{\pi}{2}) $:

$ f(-\frac{\pi}{2}) = 2*(-\frac{\pi}{2})^2 - \pi*(-\frac{\pi}{2}) + \frac{\pi^2}{4} $.

Compute each term:

First term: $ 2*(\pi^2/4) = \pi^2/2 $.

Second term: $ -\pi*(-\frac{\pi}{2}) = \pi^2/2 $.

Third term: $ \pi^2/4 $.

Adding them up: $ \pi^2/2 + \pi^2/2 + \pi^2/4 = ( \pi^2 + \pi^2 + \pi^2/2 ) / 2? Wait, let me compute step by step.

Wait:

First term is $ \pi^2/2 $, second term is $ \pi^2/2 $, third term is $ \pi^2/4 $. So altogether:

$ \pi^2/2 + \pi^2/2 = \pi^2 $. Then adding $ \pi^2/4 $, total is $ \pi^2 + \pi^2/4 = (5/4)\pi^2 $.

Wait, no: Wait, $ \pi^2/2 + \pi^2/2 = \pi^2 $, then adding $ \pi^2/4 $ gives $ \pi^2 + \pi^2/4 = (4\pi^2 + \pi^2)/4 = 5\pi^2/4 $. Yes.

Similarly, compute $ f(\frac{\pi}{2}) $:

$ f(\frac{\pi}{2}) = 2*(\frac{\pi}{2})^2 - \pi*(\frac{\pi}{2}) + \frac{\pi^2}{4} $.

Compute each term:

First term: $ 2*(\pi^2/4) = \pi^2/2 $.

Second term: $ -\pi*(\pi/2) = -\pi^2/2 $.

Third term: $ \pi^2/4 $.

Adding them together:

$ \pi^2/2 - \pi^2/2 + \pi^2/4 = 0 + \pi^2/4 = \pi^2/4 $.

Wait, so $ f(\frac{\pi}{2}) = \pi^2/4 $, and $ f(-\frac{\pi}{2}) = 5\pi^2/4 $. But hold on, since the parabola opens upwards, the minimum is at $ y = \pi/4 $, so the maximum value of the function is at the endpoint with larger value. However, since $ f(-\frac{\pi}{2}) = 5\pi^2/4 $ and $ f(\frac{\pi}{2}) = \pi^2/4 $, then the maximum value is $ 5\pi^2/4 $.

Wait, but hold on. Wait, the original function was $ f(y) = 2y^2 - \pi y + \frac{\pi^2}{4} $. So when y increases, depending on the coefficient of y, but since the parabola opens upwards, the maximum on the interval will be at whichever endpoint is further from the vertex.

But since the vertex is at y = pi/4, which is in the positive side. So when moving to the left (negative y), the function would increase? Let's check with the values: at y = -pi/2, which is further left, we have a larger value than at y = pi/2. So that seems correct.

Therefore, the range of f(y) is [pi^2/8, 5pi^2/4].

But wait, let me confirm with another way. Let me check for y in [-pi/2, pi/2], so the function f(y) is 2y² - pi y + pi²/4.

Alternatively, maybe I can write this quadratic in terms of completing the square or something else. Let me try that.

Let me write f(y) = 2y² - pi y + pi²/4.

Factor out 2 from the first two terms:

= 2(y² - (pi/2)y) + pi²/4.

Now, complete the square inside the parentheses:

Take y² - (pi/2)y. To complete the square, take half of (pi/2), which is pi/4, square it, which is pi²/16.

So:

= 2[ y² - (pi/2)y + pi²/16 - pi²/16 ] + pi²/4

= 2[ (y - pi/4)^2 - pi²/16 ] + pi²/4

= 2(y - pi/4)^2 - 2*(pi²/16) + pi²/4

Simplify:

= 2(y - pi/4)^2 - pi²/8 + pi²/4

= 2(y - pi/4)^2 + pi²/8.

Thus, f(y) = 2(y - pi/4)^2 + pi²/8.

Therefore, this is a parabola opening upwards with vertex at (pi/4, pi²/8). Therefore, the minimum value is pi²/8, and the maximum occurs at the endpoints of the interval. Since the parabola opens upwards, as we move away from the vertex, the function increases. So, the maximum will be at whichever endpoint is further from y = pi/4.

Since y ranges from -pi/2 to pi/2, let's compute the distance from pi/4:

Distance from pi/4 to pi/2 is pi/2 - pi/4 = pi/4.

Distance from pi/4 to -pi/2 is pi/2 - (-pi/2) = pi? Wait, no. Wait, distance is absolute value. Let me think.

Wait, the distance between y = pi/4 and y = -pi/2 is |pi/4 - (-pi/2)| = |pi/4 + pi/2| = |3pi/4| = 3pi/4.

Whereas distance between pi/4 and pi/2 is |pi/2 - pi/4| = pi/4. Therefore, the endpoint at -pi/2 is further away from the vertex, so the maximum value occurs at y = -pi/2.

Therefore, the maximum value is f(-pi/2) = 5pi²/4, as calculated before, and the minimum is pi²/8. Hence, the range of f(y) is [pi²/8, 5pi²/4].

But wait, let me verify with another point. Let me check the value at y = 0. If y = 0, then x = sin(0) = 0. Let's compute f(0):

Original function f(x) is (arccos 0)^2 + (arcsin 0)^2. arccos(0) is pi/2, arcsin(0) is 0. Therefore, f(0) = (pi/2)^2 + 0 = pi²/4. Which is equal to the value we found for y = pi/2 (since when y = 0, x = 0, and in our substitution, y is arcsin x. Wait, actually, when y = 0, that corresponds to x = 0, and in that case, the value is pi²/4. But in our earlier calculation, when we computed f(pi/2) which was y = pi/2, but that was for x = 1. Wait, but when y = pi/2, then x = sin(pi/2) = 1, which is correct. Similarly, y = -pi/2 corresponds to x = -1.

Therefore, our values for y at the endpoints correspond to x = 1 and x = -1. So, for example, when x = 1, arcsin(1) = pi/2, arccos(1) = 0, so f(1) = 0² + (pi/2)^2 = pi²/4. Similarly, when x = -1, arcsin(-1) = -pi/2, arccos(-1) = pi, so f(-1) = pi² + (-pi/2)^2 = pi² + pi²/4 = 5pi²/4, which matches our previous calculation. So that seems correct.

Therefore, the range is from pi²/8 to 5pi²/4. But let me check if there are any other critical points or if maybe I made a mistake in the substitution.

Wait, but another way to check is to consider specific values. For instance, let me check x = 0. As above, f(0) = (pi/2)^2 + 0 = pi²/4 ≈ 2.467. The minimum value is pi²/8 ≈ 1.234. Let me check another value. For example, take x = 1/√2 ≈ 0.707. Then arcsin(1/√2) = pi/4, arccos(1/√2) = pi/4. Therefore, f(1/√2) = (pi/4)^2 + (pi/4)^2 = 2*(pi²/16) = pi²/8 ≈ 1.234, which matches our earlier result for the minimum. So that seems correct. Therefore, that is indeed the minimum.

Similarly, check x = -1, which gives f(-1) = 5pi²/4 ≈ 12.337. Wait, pi² is about 9.8696, so 5pi²/4 is about 12.337. And pi²/4 is about 2.467, which is less than that. So yes, our range from pi²/8 to 5pi²/4 seems to fit.

But let me think again. The problem says "Find the range of f(x)". So, based on the substitution and analysis, since y is between -pi/2 and pi/2, and we found that the function f(y) reaches its minimum at pi²/8 and maximum at 5pi²/4, so the range is [pi²/8, 5pi²/4].

But let me check another value. Let me pick x = 1/2. Then arcsin(1/2) = pi/6, arccos(1/2) = pi/3. Then f(x) is (pi/3)^2 + (pi/6)^2 = pi²/9 + pi²/36 = (4pi² + pi²)/36 = 5pi²/36 ≈ 1.374. Which is between pi²/8 ≈ 1.234 and 5pi²/4 ≈ 12.337, so that seems okay.

Another check, x = sqrt(2)/2 ≈ 0.707, which we already did, gives pi²/8. What about x approaching 1? Let me take x approaching 1, so arcsin x approaches pi/2, arccos x approaches 0. So f(x) approaches (0)^2 + (pi/2)^2 = pi²/4 ≈ 2.467, which is between pi²/8 and 5pi²/4.

Similarly, when x approaches -1, arcsin x approaches -pi/2 and arccos x approaches pi. Therefore, f(x) approaches pi² + ( -pi/2)^2 = pi² + pi²/4 = 5pi²/4, which matches.

Therefore, all these checks confirm that the minimum value is pi²/8 and the maximum is 5pi²/4. Therefore, the range is [pi²/8, 5pi²/4].

Wait, but let me think again. The original function is f(x) = (arccos x)^2 + (arcsin x)^2. Since arccos x and arcsin x are both defined on [-1, 1], and we used the identity to convert it into a function of y = arcsin x, which leads us to the quadratic. So all steps seem valid.

But just to confirm, is there any possible mistake in the substitution?

Original substitution:

Since arccos x = pi/2 - arcsin x, then f(x) = (pi/2 - y)^2 + y² where y = arcsin x. Then expanding gives 2y² - pi y + pi²/4. Then analyzing that quadratic, which seems okay.

Therefore, unless I made an error in completing the square or calculating the vertex, which I think I didn't, the conclusion holds.

Therefore, the range of f(x) is from pi squared over 8 to 5 pi squared over 4.

But let me check once more for another value. Let me take y = pi/2, which gives f(y) = 5pi²/4. Wait, when y = pi/2, which is arcsin x = pi/2, which implies x = 1. Therefore, arccos(1) = 0, so f(1) = 0² + (pi/2)^2 = pi²/4. Wait, hold on! Wait, there is a contradiction here. Wait, when I calculated earlier, I had for f(-pi/2) = 5pi²/4, but when y = pi/2, which corresponds to x = 1, then f(1) should be pi²/4. However, when I calculated f(pi/2) in terms of the quadratic, I found pi²/4. So why did we have f(-pi/2) = 5pi²/4? Because y = -pi/2 corresponds to x = -1, which leads to arccos(-1) = pi and arcsin(-1) = -pi/2. Therefore, f(-1) = pi² + (-pi/2)^2 = pi² + pi²/4 = 5pi²/4. So that is correct. Therefore, there's no contradiction here. When y = pi/2, which is arcsin(1) = pi/2, then the value is pi²/4, which is less than 5pi²/4. So the maximum is indeed at y = -pi/2.

Therefore, the conclusion is correct. Therefore, the range of f(x) is [π²/8, 5π²/4].

But just to make sure, let me check if there's another way to approach this problem, maybe using calculus on the original function. Let me try that.

Consider the original function f(x) = (arccos x)^2 + (arcsin x)^2. Since we know that arccos x + arcsin x = π/2, perhaps we can use calculus to find critical points. Let me try taking derivative of f(x) with respect to x.

First, compute f'(x):

Let me denote:

f(x) = (arccos x)^2 + (arcsin x)^2.

Therefore,

f'(x) = 2 arccos x * d/dx(arccos x) + 2 arcsin x * d/dx(arcsin x).

We know that derivative of arccos x is -1 / sqrt(1 - x²), and derivative of arcsin x is 1 / sqrt(1 - x²).

Therefore:

f'(x) = 2 arccos x * (-1 / sqrt(1 - x²)) + 2 arcsin x * (1 / sqrt(1 - x²)).

Factor out 2 / sqrt(1 - x²):

f'(x) = [2 / sqrt(1 - x²)] [ - arccos x + arcsin x ].

But we know that arcsin x + arccos x = π/2, so arcsin x = π/2 - arccos x. Therefore:

- arccos x + arcsin x = - arccos x + π/2 - arccos x = π/2 - 2 arccos x.

Therefore, f'(x) = [2 / sqrt(1 - x²)] [ π/2 - 2 arccos x ].

Set derivative equal to zero to find critical points:

[2 / sqrt(1 - x²)] [ π/2 - 2 arccos x ] = 0.

Since sqrt(1 - x²) is never zero for x in (-1, 1), the only critical points occur when π/2 - 2 arccos x = 0.

Solve π/2 - 2 arccos x = 0 => 2 arccos x = π/2 => arccos x = π/4 => x = cos(π/4) = √2/2 ≈ 0.707.

Therefore, the only critical point in (-1, 1) is at x = √2/2.

Now, to find the range, we need to evaluate f(x) at critical points and endpoints. The endpoints are x = -1 and x = 1.

We already calculated f(-1) = 5π²/4, f(1) = π²/4, and f(√2/2) = π²/8.

Therefore, the minimum is at x = √2/2 with value π²/8, and the maximum is at x = -1 with value 5π²/4. Therefore, the range is [π²/8, 5π²/4].

This matches our previous conclusion, so that's reassuring. Therefore, I can be confident that the range of f(x) is from π²/8 to 5π²/4.

Just to make sure, let me check if there are any other critical points. The derivative is zero only at x = √2/2, and we also need to check the behavior at the endpoints. Since the function is continuous on [-1, 1], and differentiable on (-1, 1), then the extrema must occur at critical points or endpoints. Therefore, we have checked all possibilities.

Thus, the range is indeed [π²/8, 5π²/4].

**Final Answer**
The range of $ f(x) $ is \boxed{\left[ \dfrac{\pi^2}{8}, \dfrac{5\pi^2}{4} \right]}.

To find the **range** of the function

$$
f(x) = (\arccos x)^2 + (\arcsin x)^2,
$$

we begin by recalling the key identity:

$$
\arccos x + \arcsin x = \frac{\pi}{2} \quad \text{for all } x \in [-1, 1].
$$

Let us define:

$$
y = \arcsin x.
$$

Then, using the above identity:

$$
\arccos x = \frac{\pi}{2} - y.
$$

Therefore, the function becomes:

$$
f(x) = \left(\frac{\pi}{2} - y\right)^2 + y^2.
$$

Expanding this expression:

$$
f(x) = \left(\frac{\pi^2}{4} - \pi y + y^2\right) + y^2 = \frac{\pi^2}{4} - \pi y + 2y^2.
$$

So we are now evaluating the function:

$$
f(y) = 2y^2 - \pi y + \frac{\pi^2}{4},
$$

with the constraint that $ y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] $ (since $ \arcsin x $ maps $ x \in [-1, 1] $ to that range).

---

### Step 1: Determine Extrema of the Quadratic Function

This is a **quadratic** function in $ y $, and the coefficient of $ y^2 $ is positive (2), so the parabola opens **upwards**. This means:

- **Minimum** occurs at the **vertex**.
- **Maximum** occurs at an endpoint.

The vertex of a quadratic $ ay^2 + by + c $ is at:

$$
y = -\frac{b}{2a}.
$$

Here, $ a = 2 $, $ b = -\pi $, so the vertex is at:

$$
y = \frac{\pi}{4}.
$$

We now evaluate $ f(y) $ at this point and at the endpoints $ y = -\frac{\pi}{2} $ and $ y = \frac{\pi}{2} $.

---

### Step 2: Compute Values of $ f(y) $

#### At $ y = \frac{\pi}{4} $:

$$
f\left(\frac{\pi}{4}\right) = 2\left(\frac{\pi}{4}\right)^2 - \pi\left(\frac{\pi}{4}\right) + \frac{\pi^2}{4}
= 2\cdot \frac{\pi^2}{16} - \frac{\pi^2}{4} + \frac{\pi^2}{4} = \frac{\pi^2}{8}.
$$

#### At $ y = -\frac{\pi}{2} $:

$$
f\left(-\frac{\pi}{2}\right) = 2\left(\frac{\pi^2}{4}\right) + \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{\pi^2}{2} + \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{5\pi^2}{4}.
$$

#### At $ y = \frac{\pi}{2} $:

$$
f\left(\frac{\pi}{2}\right) = 2\cdot \frac{\pi^2}{4} - \pi\cdot \frac{\pi}{2} + \frac{\pi^2}{4}
= \frac{\pi^2}{2} - \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{\pi^2}{4}.
$$

---

### Step 3: Analyze the Range

- The **minimum** value is $ \frac{\pi^2}{8} $ — achieved at $ y = \frac{\pi}{4} $.
- The **maximum** value is $ \frac{5\pi^2}{4} $ — achieved at $ y = -\frac{\pi}{2} $ (i.e., $ x = -1 $).

Thus, the **range** of $ f(x) $ is:

$$
\left[\frac{\pi^2}{8}, \frac{5\pi^2}{4}\right].
$$

---

### Final Answer

$$
\boxed{\left[ \dfrac{\pi^2}{8}, \dfrac{5\pi^2}{4} \right]}
$$

[Assistant]
Explanation:
Okay, so I need to find the range of the function $ f(x) = (\arccos x)^2 + (\arcsin x)^2 $. Hmm, let me start by recalling some basics about inverse trigonometric functions.

First, I know that the domains of both $ \arccos x $ and $ \arcsin x $ are $ [-1, 1] $. So, the domain of $ f(x) $ is also $ [-1, 1] $. That's straightforward. Now, I need to figure out the range of this function.

I remember that there's a relationship between $ \arccos x $ and $ \arcsin x $. Specifically, $ \arccos x + \arcsin x = \frac{\pi}{2} $ for all $ x $ in $ [-1, 1] $. Let me verify that... If I take $ \arccos x + \arcsin x $, let me denote $ \theta = \arcsin x $. Then $ \sin \theta = x $, and $ \theta $ is between $ -\frac{\pi}{2} $ and $ \frac{\pi}{2} $, but since $ \arcsin x $ is defined to return values in $ [-\frac{\pi}{2}, \frac{\pi}{2}] $. Wait, but $ \arccos x $ is in $ [0, \pi] $. So adding them together...

Alternatively, maybe I can use the identity. Let me think, if $ \theta = \arcsin x $, then $ \arccos x = \frac{\pi}{2} - \theta $. Because $ \cos\left(\frac{\pi}{2} - \theta\right) = \sin \theta = x $. Yes, so $ \arccos x = \frac{\pi}{2} - \arcsin x $. Therefore, their sum is $ \frac{\pi}{2} $.

So that means $ (\arccos x)^2 + (\arcsin x)^2 $ can be rewritten using this identity. Let me try substituting $ \arccos x = \frac{\pi}{2} - \arcsin x $ into the function.

Let me denote $ y = \arcsin x $. Then $ \arccos x = \frac{\pi}{2} - y $, so the function becomes:

$ f(x) = \left( \frac{\pi}{2} - y \right)^2 + y^2 $.

Since $ y = \arcsin x $, and $ x \in [-1, 1] $, then $ y \in [-\frac{\pi}{2}, \frac{\pi}{2}] $. However, $ \arcsin x $ actually returns values in $ [-\frac{\pi}{2}, \frac{\pi}{2}] $, but since $ \arccos x $ is in $ [0, \pi] $, but we already used that relationship. Wait, but if $ y $ is in $ [-\frac{\pi}{2}, \frac{\pi}{2}] $, then $ \frac{\pi}{2} - y $ would be in $ [\frac{\pi}{2} - \frac{\pi}{2}, \frac{\pi}{2} - (-\frac{\pi}{2})] = [0, \pi] $, which matches the range of $ \arccos x $. So that substitution is valid.

So, substituting, we have:

$ f(x) = \left( \frac{\pi}{2} - y \right)^2 + y^2 $.

Let me expand that expression:

First, square the terms:

$ \left( \frac{\pi}{2} - y \right)^2 = \left( \frac{\pi}{2} \right)^2 - \pi y + y^2 $.

Therefore, adding $ y^2 $:

$ f(x) = \left( \frac{\pi^2}{4} - \pi y + y^2 \right) + y^2 = \frac{\pi^2}{4} - \pi y + 2y^2 $.

So, simplifying, the function becomes $ f(y) = 2y^2 - \pi y + \frac{\pi^2}{4} $, where $ y \in [-\frac{\pi}{2}, \frac{\pi}{2}] $.

Now, since this is a quadratic in terms of $ y $, I can analyze its minimum and maximum values over the interval $ y \in [-\frac{\pi}{2}, \frac{\pi}{2}] $.

Let me write the quadratic again: $ f(y) = 2y^2 - \pi y + \frac{\pi^2}{4} $. To find its extrema, I can find its derivative and critical points, or use vertex formula. Since it's a quadratic, we know it's a parabola. The coefficient of $ y^2 $ is 2, which is positive, so the parabola opens upwards. Therefore, it has a minimum at its vertex and maximum at one of the endpoints of the interval.

So, first, let's find the vertex. The vertex of a quadratic $ ay^2 + by + c $ is at $ y = -\frac{b}{2a} $.

Here, $ a = 2 $, $ b = -\pi $, so the vertex is at:

$ y = -(-\pi)/(2*2) = \pi/4 $.

So the minimum occurs at $ y = \pi/4 $. Now, I need to check if this point is within the interval $ [-\frac{\pi}{2}, \frac{\pi}{2}] $. Well, $ \pi/4 $ is approximately 0.785 radians, and $ \pi/2 $ is approximately 1.5708 radians. So yes, $ \pi/4 $ is within the interval. Therefore, the minimum value of $ f(y) $ is at $ y = \pi/4 $, and the maximum occurs at one of the endpoints, either $ y = -\frac{\pi}{2} $ or $ y = \frac{\pi}{2} $.

Therefore, to find the range of $ f(x) $, I need to compute $ f(\pi/4) $, $ f(-\frac{\pi}{2}) $, and $ f(\frac{\pi}{2}) $, then determine the minimum and maximum values.

Let me compute each of these.

First, compute $ f(\pi/4) $:

$ f(\pi/4) = 2(\pi/4)^2 - \pi(\pi/4) + \frac{\pi^2}{4} $.

Let me compute each term:

First term: $ 2*(\pi^2/16) = \pi^2/8 $.

Second term: $ - \pi*(\pi/4) = - \pi^2/4 $.

Third term: $ \pi^2/4 $.

Adding them together:

$ \pi^2/8 - \pi^2/4 + \pi^2/4 = \pi^2/8 $. The $ -\pi^2/4 $ and $ +\pi^2/4 $ cancel out, leaving $ \pi^2/8 $.

So, the minimum value is $ \pi^2/8 $.

Now, compute $ f(-\frac{\pi}{2}) $:

$ f(-\frac{\pi}{2}) = 2*(-\frac{\pi}{2})^2 - \pi*(-\frac{\pi}{2}) + \frac{\pi^2}{4} $.

Compute each term:

First term: $ 2*(\pi^2/4) = \pi^2/2 $.

Second term: $ -\pi*(-\frac{\pi}{2}) = \pi^2/2 $.

Third term: $ \pi^2/4 $.

Adding them up: $ \pi^2/2 + \pi^2/2 + \pi^2/4 = ( \pi^2 + \pi^2 + \pi^2/2 ) / 2? Wait, let me compute step by step.

Wait:

First term is $ \pi^2/2 $, second term is $ \pi^2/2 $, third term is $ \pi^2/4 $. So altogether:

$ \pi^2/2 + \pi^2/2 = \pi^2 $. Then adding $ \pi^2/4 $, total is $ \pi^2 + \pi^2/4 = (5/4)\pi^2 $.

Wait, no: Wait, $ \pi^2/2 + \pi^2/2 = \pi^2 $, then adding $ \pi^2/4 $ gives $ \pi^2 + \pi^2/4 = (4\pi^2 + \pi^2)/4 = 5\pi^2/4 $. Yes.

Similarly, compute $ f(\frac{\pi}{2}) $:

$ f(\frac{\pi}{2}) = 2*(\frac{\pi}{2})^2 - \pi*(\frac{\pi}{2}) + \frac{\pi^2}{4} $.

Compute each term:

First term: $ 2*(\pi^2/4) = \pi^2/2 $.

Second term: $ -\pi*(\pi/2) = -\pi^2/2 $.

Third term: $ \pi^2/4 $.

Adding them together:

$ \pi^2/2 - \pi^2/2 + \pi^2/4 = 0 + \pi^2/4 = \pi^2/4 $.

Wait, so $ f(\frac{\pi}{2}) = \pi^2/4 $, and $ f(-\frac{\pi}{2}) = 5\pi^2/4 $. But hold on, since the parabola opens upwards, the minimum is at $ y = \pi/4 $, so the maximum value of the function is at the endpoint with larger value. However, since $ f(-\frac{\pi}{2}) = 5\pi^2/4 $ and $ f(\frac{\pi}{2}) = \pi^2/4 $, then the maximum value is $ 5\pi^2/4 $.

Wait, but hold on. Wait, the original function was $ f(y) = 2y^2 - \pi y + \frac{\pi^2}{4} $. So when y increases, depending on the coefficient of y, but since the parabola opens upwards, the maximum on the interval will be at whichever endpoint is further from the vertex.

But since the vertex is at y = pi/4, which is in the positive side. So when moving to the left (negative y), the function would increase? Let's check with the values: at y = -pi/2, which is further left, we have a larger value than at y = pi/2. So that seems correct.

Therefore, the range of f(y) is [pi^2/8, 5pi^2/4].

But wait, let me confirm with another way. Let me check for y in [-pi/2, pi/2], so the function f(y) is 2y² - pi y + pi²/4.

Alternatively, maybe I can write this quadratic in terms of completing the square or something else. Let me try that.

Let me write f(y) = 2y² - pi y + pi²/4.

Factor out 2 from the first two terms:

= 2(y² - (pi/2)y) + pi²/4.

Now, complete the square inside the parentheses:

Take y² - (pi/2)y. To complete the square, take half of (pi/2), which is pi/4, square it, which is pi²/16.

So:

= 2[ y² - (pi/2)y + pi²/16 - pi²/16 ] + pi²/4

= 2[ (y - pi/4)^2 - pi²/16 ] + pi²/4

= 2(y - pi/4)^2 - 2*(pi²/16) + pi²/4

Simplify:

= 2(y - pi/4)^2 - pi²/8 + pi²/4

= 2(y - pi/4)^2 + pi²/8.

Thus, f(y) = 2(y - pi/4)^2 + pi²/8.

Therefore, this is a parabola opening upwards with vertex at (pi/4, pi²/8). Therefore, the minimum value is pi²/8, and the maximum occurs at the endpoints of the interval. Since the parabola opens upwards, as we move away from the vertex, the function increases. So, the maximum will be at whichever endpoint is further from y = pi/4.

Since y ranges from -pi/2 to pi/2, let's compute the distance from pi/4:

Distance from pi/4 to pi/2 is pi/2 - pi/4 = pi/4.

Distance from pi/4 to -pi/2 is pi/2 - (-pi/2) = pi? Wait, no. Wait, distance is absolute value. Let me think.

Wait, the distance between y = pi/4 and y = -pi/2 is |pi/4 - (-pi/2)| = |pi/4 + pi/2| = |3pi/4| = 3pi/4.

Whereas distance between pi/4 and pi/2 is |pi/2 - pi/4| = pi/4. Therefore, the endpoint at -pi/2 is further away from the vertex, so the maximum value occurs at y = -pi/2.

Therefore, the maximum value is f(-pi/2) = 5pi²/4, as calculated before, and the minimum is pi²/8. Hence, the range of f(y) is [pi²/8, 5pi²/4].

But wait, let me verify with another point. Let me check the value at y = 0. If y = 0, then x = sin(0) = 0. Let's compute f(0):

Original function f(x) is (arccos 0)^2 + (arcsin 0)^2. arccos(0) is pi/2, arcsin(0) is 0. Therefore, f(0) = (pi/2)^2 + 0 = pi²/4. Which is equal to the value we found for y = pi/2 (since when y = 0, x = 0, and in our substitution, y is arcsin x. Wait, actually, when y = 0, that corresponds to x = 0, and in that case, the value is pi²/4. But in our earlier calculation, when we computed f(pi/2) which was y = pi/2, but that was for x = 1. Wait, but when y = pi/2, then x = sin(pi/2) = 1, which is correct. Similarly, y = -pi/2 corresponds to x = -1.

Therefore, our values for y at the endpoints correspond to x = 1 and x = -1. So, for example, when x = 1, arcsin(1) = pi/2, arccos(1) = 0, so f(1) = 0² + (pi/2)^2 = pi²/4. Similarly, when x = -1, arcsin(-1) = -pi/2, arccos(-1) = pi, so f(-1) = pi² + (-pi/2)^2 = pi² + pi²/4 = 5pi²/4, which matches our previous calculation. So that seems correct.

Therefore, the range is from pi²/8 to 5pi²/4. But let me check if there are any other critical points or if maybe I made a mistake in the substitution.

Wait, but another way to check is to consider specific values. For instance, let me check x = 0. As above, f(0) = (pi/2)^2 + 0 = pi²/4 ≈ 2.467. The minimum value is pi²/8 ≈ 1.234. Let me check another value. For example, take x = 1/√2 ≈ 0.707. Then arcsin(1/√2) = pi/4, arccos(1/√2) = pi/4. Therefore, f(1/√2) = (pi/4)^2 + (pi/4)^2 = 2*(pi²/16) = pi²/8 ≈ 1.234, which matches our earlier result for the minimum. So that seems correct. Therefore, that is indeed the minimum.

Similarly, check x = -1, which gives f(-1) = 5pi²/4 ≈ 12.337. Wait, pi² is about 9.8696, so 5pi²/4 is about 12.337. And pi²/4 is about 2.467, which is less than that. So yes, our range from pi²/8 to 5pi²/4 seems to fit.

But let me think again. The problem says "Find the range of f(x)". So, based on the substitution and analysis, since y is between -pi/2 and pi/2, and we found that the function f(y) reaches its minimum at pi²/8 and maximum at 5pi²/4, so the range is [pi²/8, 5pi²/4].

But let me check another value. Let me pick x = 1/2. Then arcsin(1/2) = pi/6, arccos(1/2) = pi/3. Then f(x) is (pi/3)^2 + (pi/6)^2 = pi²/9 + pi²/36 = (4pi² + pi²)/36 = 5pi²/36 ≈ 1.374. Which is between pi²/8 ≈ 1.234 and 5pi²/4 ≈ 12.337, so that seems okay.

Another check, x = sqrt(2)/2 ≈ 0.707, which we already did, gives pi²/8. What about x approaching 1? Let me take x approaching 1, so arcsin x approaches pi/2, arccos x approaches 0. So f(x) approaches (0)^2 + (pi/2)^2 = pi²/4 ≈ 2.467, which is between pi²/8 and 5pi²/4.

Similarly, when x approaches -1, arcsin x approaches -pi/2 and arccos x approaches pi. Therefore, f(x) approaches pi² + ( -pi/2)^2 = pi² + pi²/4 = 5pi²/4, which matches.

Therefore, all these checks confirm that the minimum value is pi²/8 and the maximum is 5pi²/4. Therefore, the range is [pi²/8, 5pi²/4].

Wait, but let me think again. The original function is f(x) = (arccos x)^2 + (arcsin x)^2. Since arccos x and arcsin x are both defined on [-1, 1], and we used the identity to convert it into a function of y = arcsin x, which leads us to the quadratic. So all steps seem valid.

But just to confirm, is there any possible mistake in the substitution?

Original substitution:

Since arccos x = pi/2 - arcsin x, then f(x) = (pi/2 - y)^2 + y² where y = arcsin x. Then expanding gives 2y² - pi y + pi²/4. Then analyzing that quadratic, which seems okay.

Therefore, unless I made an error in completing the square or calculating the vertex, which I think I didn't, the conclusion holds.

Therefore, the range of f(x) is from pi squared over 8 to 5 pi squared over 4.

But let me check once more for another value. Let me take y = pi/2, which gives f(y) = 5pi²/4. Wait, when y = pi/2, which is arcsin x = pi/2, which implies x = 1. Therefore, arccos(1) = 0, so f(1) = 0² + (pi/2)^2 = pi²/4. Wait, hold on! Wait, there is a contradiction here. Wait, when I calculated earlier, I had for f(-pi/2) = 5pi²/4, but when y = pi/2, which corresponds to x = 1, then f(1) should be pi²/4. However, when I calculated f(pi/2) in terms of the quadratic, I found pi²/4. So why did we have f(-pi/2) = 5pi²/4? Because y = -pi/2 corresponds to x = -1, which leads to arccos(-1) = pi and arcsin(-1) = -pi/2. Therefore, f(-1) = pi² + (-pi/2)^2 = pi² + pi²/4 = 5pi²/4. So that is correct. Therefore, there's no contradiction here. When y = pi/2, which is arcsin(1) = pi/2, then the value is pi²/4, which is less than 5pi²/4. So the maximum is indeed at y = -pi/2.

Therefore, the conclusion is correct. Therefore, the range of f(x) is [π²/8, 5π²/4].

But just to make sure, let me check if there's another way to approach this problem, maybe using calculus on the original function. Let me try that.

Consider the original function f(x) = (arccos x)^2 + (arcsin x)^2. Since we know that arccos x + arcsin x = π/2, perhaps we can use calculus to find critical points. Let me try taking derivative of f(x) with respect to x.

First, compute f'(x):

Let me denote:

f(x) = (arccos x)^2 + (arcsin x)^2.

Therefore,

f'(x) = 2 arccos x * d/dx(arccos x) + 2 arcsin x * d/dx(arcsin x).

We know that derivative of arccos x is -1 / sqrt(1 - x²), and derivative of arcsin x is 1 / sqrt(1 - x²).

Therefore:

f'(x) = 2 arccos x * (-1 / sqrt(1 - x²)) + 2 arcsin x * (1 / sqrt(1 - x²)).

Factor out 2 / sqrt(1 - x²):

f'(x) = [2 / sqrt(1 - x²)] [ - arccos x + arcsin x ].

But we know that arcsin x + arccos x = π/2, so arcsin x = π/2 - arccos x. Therefore:

- arccos x + arcsin x = - arccos x + π/2 - arccos x = π/2 - 2 arccos x.

Therefore, f'(x) = [2 / sqrt(1 - x²)] [ π/2 - 2 arccos x ].

Set derivative equal to zero to find critical points:

[2 / sqrt(1 - x²)] [ π/2 - 2 arccos x ] = 0.

Since sqrt(1 - x²) is never zero for x in (-1, 1), the only critical points occur when π/2 - 2 arccos x = 0.

Solve π/2 - 2 arccos x = 0 => 2 arccos x = π/2 => arccos x = π/4 => x = cos(π/4) = √2/2 ≈ 0.707.

Therefore, the only critical point in (-1, 1) is at x = √2/2.

Now, to find the range, we need to evaluate f(x) at critical points and endpoints. The endpoints are x = -1 and x = 1.

We already calculated f(-1) = 5π²/4, f(1) = π²/4, and f(√2/2) = π²/8.

Therefore, the minimum is at x = √2/2 with value π²/8, and the maximum is at x = -1 with value 5π²/4. Therefore, the range is [π²/8, 5π²/4].

This matches our previous conclusion, so that's reassuring. Therefore, I can be confident that the range of f(x) is from π²/8 to 5π²/4.

Just to make sure, let me check if there are any other critical points. The derivative is zero only at x = √2/2, and we also need to check the behavior at the endpoints. Since the function is continuous on [-1, 1], and differentiable on (-1, 1), then the extrema must occur at critical points or endpoints. Therefore, we have checked all possibilities.

Thus, the range is indeed [π²/8, 5π²/4].

**Final Answer**
The range of $ f(x) $ is \boxed{\left[ \dfrac{\pi^2}{8}, \dfrac{5\pi^2}{4} \right]}.

To find the **range** of the function

$$
f(x) = (\arccos x)^2 + (\arcsin x)^2,
$$

we begin by recalling the key identity:

$$
\arccos x + \arcsin x = \frac{\pi}{2} \quad \text{for all } x \in [-1, 1].
$$

Let us define:

$$
y = \arcsin x.
$$

Then, using the above identity:

$$
\arccos x = \frac{\pi}{2} - y.
$$

Therefore, the function becomes:

$$
f(x) = \left(\frac{\pi}{2} - y\right)^2 + y^2.
$$

Expanding this expression:

$$
f(x) = \left(\frac{\pi^2}{4} - \pi y + y^2\right) + y^2 = \frac{\pi^2}{4} - \pi y + 2y^2.
$$

So we are now evaluating the function:

$$
f(y) = 2y^2 - \pi y + \frac{\pi^2}{4},
$$

with the constraint that $ y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] $ (since $ \arcsin x $ maps $ x \in [-1, 1] $ to that range).

---

### Step 1: Determine Extrema of the Quadratic Function

This is a **quadratic** function in $ y $, and the coefficient of $ y^2 $ is positive (2), so the parabola opens **upwards**. This means:

- **Minimum** occurs at the **vertex**.
- **Maximum** occurs at an endpoint.

The vertex of a quadratic $ ay^2 + by + c $ is at:

$$
y = -\frac{b}{2a}.
$$

Here, $ a = 2 $, $ b = -\pi $, so the vertex is at:

$$
y = \frac{\pi}{4}.
$$

We now evaluate $ f(y) $ at this point and at the endpoints $ y = -\frac{\pi}{2} $ and $ y = \frac{\pi}{2} $.

---

### Step 2: Compute Values of $ f(y) $

#### At $ y = \frac{\pi}{4} $:

$$
f\left(\frac{\pi}{4}\right) = 2\left(\frac{\pi}{4}\right)^2 - \pi\left(\frac{\pi}{4}\right) + \frac{\pi^2}{4}
= 2\cdot \frac{\pi^2}{16} - \frac{\pi^2}{4} + \frac{\pi^2}{4} = \frac{\pi^2}{8}.
$$

#### At $ y = -\frac{\pi}{2} $:

$$
f\left(-\frac{\pi}{2}\right) = 2\left(\frac{\pi^2}{4}\right) + \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{\pi^2}{2} + \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{5\pi^2}{4}.
$$

#### At $ y = \frac{\pi}{2} $:

$$
f\left(\frac{\pi}{2}\right) = 2\cdot \frac{\pi^2}{4} - \pi\cdot \frac{\pi}{2} + \frac{\pi^2}{4}
= \frac{\pi^2}{2} - \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{\pi^2}{4}.
$$

---

### Step 3: Analyze the Range

- The **minimum** value is $ \frac{\pi^2}{8} $ — achieved at $ y = \frac{\pi}{4} $.
- The **maximum** value is $ \frac{5\pi^2}{4} $ — achieved at $ y = -\frac{\pi}{2} $ (i.e., $ x = -1 $).

Thus, the **range** of $ f(x) $ is:

$$
\left[\frac{\pi^2}{8}, \frac{5\pi^2}{4}\right].
$$

---

### Final Answer

$$
\boxed{\left[ \dfrac{\pi^2}{8}, \dfrac{5\pi^2}{4} \right]}
$$

Solution:
First, we claim that $\arccos x + \arcsin x = \frac{\pi}{2}$ for all $x \in [-1,1].$

Note that
\[\cos \left( \frac{\pi}{2} - \arcsin x \right) = \cos (\arccos x) = x.\]Furthermore, $-\frac{\pi}{2} \le \arcsin x \le \frac{\pi}{2},$ so $0 \le \frac{\pi}{2} - \arcsin x \le \pi.$  Therefore,
\[\frac{\pi}{2} - \arcsin x = \arccos x,\]so $\arccos x + \arcsin x = \frac{\pi}{2}.$

Let $\alpha = \arccos x$ and $\beta = \arcsin x,$ so $\alpha + \beta = \frac{\pi}{2}.$  Then
\begin{align*}
f(x) &= (\arccos x)^2 + (\arcsin x)^2 \\
&= \alpha^2 + \beta^2 \\
&= \left( \frac{\pi}{2} - \beta \right)^2 + \beta^2 \\
&= 2 \beta^2 - \pi \beta + \frac{\pi^2}{4} \\
&= 2 \left( \beta - \frac{\pi}{4} \right)^2 + \frac{\pi^2}{8}.
\end{align*}Since $-\frac{\pi}{2} \le \beta \le \frac{\pi}{2},$ the range of $f(x)$ is $\boxed{\left[ \frac{\pi^2}{8}, \frac{5 \pi^2}{4} \right]}.$


Explanation:
Okay, so I need to find the number of dice, n, such that when you roll them all at the same time, the probability that exactly two of them show a number other than 1 is 25/216. Hmm, let me try to break this down.

First, let me recall some probability concepts. When dealing with dice, each die has 6 sides, numbered 1 through 6. The problem says "exactly two of them show a number other than 1." So, that means out of the n dice, exactly two are not 1, and the rest (n - 2) are 1s. Wait, is that right? Let me check.

Yes, exactly two show a number other than 1, so the other n - 2 must show 1. So, the probability of this happening is calculated using the binomial probability formula. The binomial formula is P(k) = C(n, k) * p^k * (1 - p)^(n - k), where p is the probability of success on a single trial. In this case, a "success" would be rolling a number other than 1. Let me confirm.

Each die has a probability of 5/6 of showing a number other than 1, right? Because there are 5 numbers that are not 1 (2 through 6). And the probability of rolling a 1 is 1/6. So, if we're looking for exactly two dice that are not 1, then k = 2. So, substituting into the formula:

P(2) = C(n, 2) * (5/6)^2 * (1/6)^(n - 2)

And this is given to be equal to 25/216. So, we need to solve for n in the equation:

C(n, 2) * (5/6)^2 * (1/6)^(n - 2) = 25/216

Okay, let me write that down more clearly:

C(n, 2) * (25/36) * (1/6)^{n - 2} = 25/216

First, maybe I can simplify this equation. Let me note that 25/36 is (5/6)^2, and 25/216 is 25 divided by 6^3. Let me see if I can cancel some terms.

Divide both sides of the equation by 25/36:

C(n, 2) * (1/6)^{n - 2} = (25/216) / (25/36) = (25/216) * (36/25) = 36/216 = 1/6

So, simplifying, we have:

C(n, 2) * (1/6)^{n - 2} = 1/6

Hmm, so let me write that equation again:

C(n, 2) * (1/6)^{n - 2} = 1/6

Let me think about how to solve this equation for n. Since n is an integer greater than or equal to 2 (because we need exactly two dice to show non-1 numbers), we can try plugging in values of n and see which one satisfies the equation. Alternatively, maybe I can manipulate the equation algebraically.

First, let me recall that C(n, 2) is equal to n(n - 1)/2. So substituting that in:

[n(n - 1)/2] * (1/6)^{n - 2} = 1/6

Multiply both sides by 2:

n(n - 1) * (1/6)^{n - 2} = 2/6 = 1/3

So:

n(n - 1) * (1/6)^{n - 2} = 1/3

Hmm. Let me rewrite this equation as:

n(n - 1) / 6^{n - 2} = 1/3

Multiply both sides by 6^{n - 2}:

n(n - 1) = (1/3) * 6^{n - 2}

Alternatively:

n(n - 1) = 6^{n - 2} / 3

Let me note that 6^{n - 2} / 3 is equal to 6^{n - 2} * 3^{-1} = 6^{n - 2} / 3. Alternatively, since 6^{n - 2} = 6^{n} / 6^2 = 6^n / 36, so substituting back:

n(n - 1) = (6^n / 36) / 3 = 6^n / 108

Therefore:

n(n - 1) = 6^n / 108

Multiply both sides by 108:

108n(n - 1) = 6^n

Hmm. So, 6^n = 108n(n - 1). Let me note that 108 is 6^2 * 3, since 6^2 is 36 and 36*3 is 108. So, 6^n = 6^2 * 3 * n(n - 1). Therefore:

6^{n - 2} = 3n(n - 1)

So, 6^{n - 2} = 3n(n - 1). Hmm. Maybe this equation is easier to work with. Let me write that:

6^{n - 2} = 3n(n - 1)

Alternatively, divide both sides by 3:

2 * 6^{n - 2} = n(n - 1)

Wait, 6^{n - 2} divided by 3 is 6^{n - 2}/3 = 2^{n - 2} * 3^{n - 2}/3 = 2^{n - 2} * 3^{n - 3}. Maybe that's not helpful. Alternatively, let me try plugging in small integer values for n and see if I can find a solution.

Let me start with n=2. Let me check:

For n=2:

Left side: 6^{2 - 2} = 6^0 = 1

Right side: 3*2*1 = 6. 1 vs 6: Not equal.

n=3:

Left side: 6^{3 - 2} = 6^1 = 6

Right side: 3*3*2 = 18. 6 vs 18: Not equal.

n=4:

Left side: 6^{4 - 2} = 6^2 = 36

Right side: 3*4*3 = 36. Oh! 36 = 36. So, n=4 is a solution.

Wait, so n=4 works? Let me check if there are other solutions. Let me check n=5.

Left side: 6^{5 - 2} = 6^3 = 216

Right side: 3*5*4 = 60. 216 vs 60: Not equal.

n=6:

Left side: 6^{6 - 2} = 6^4 = 1296

Right side: 3*6*5 = 90. 1296 vs 90: Nope.

What about n=1? Wait, n must be at least 2 because we need exactly two dice to show non-1 numbers. If n=1, you can't have exactly two dice. So n=2 is the minimum. So n=4 is a solution. Let me check n=0? Not relevant. Negative numbers? No.

But let me check if there are other solutions. For example, n=5 gives left side 216 vs right side 60. So left side increases exponentially, while the right side is quadratic. So, for n=4, they are equal. For n>4, left side grows much faster. For n=3, left side is 6 vs right side 18. So for n=3, right side is bigger. So, n=4 is the only integer where they are equal?

Wait, is there another value? Let me check n=5 again. 6^3=216, right side 60. So left is bigger. What about n=4? 36 vs 36. For n=3, 6 vs 18. For n=2, 1 vs 6. So for n=4, it works, and for n>4, left side grows faster. So, only n=4?

But let me check if maybe there's another value where maybe the equation holds. Let me check n= something else? Like n= something smaller? But n has to be at least 2. For n=2, left side 1 vs 6. No. For n=3, 6 vs 18. No. So only n=4? So that suggests that n=4 is the answer. Let me check my steps again to make sure I didn't make a mistake.

Original probability was given as 25/216. Let me check if for n=4, the probability is indeed 25/216. Let me compute it again.

For n=4, probability that exactly two show numbers other than 1. So, number of ways: C(4, 2). Then, probability for each such case: (5/6)^2*(1/6)^{4-2} = (5/6)^2*(1/6)^2. Then, multiply by C(4, 2).

So, C(4, 2) is 6. Then 6*(25/36)*(1/36) ?

Wait, no. Wait: (5/6)^2 is 25/36, and (1/6)^{4-2} is (1/6)^2 = 1/36. Then multiply them together: 25/36 * 1/36 = 25 / 1296. Multiply by 6: 6 * 25 / 1296 = 150 / 1296. Simplify this fraction: divide numerator and denominator by 6: 25 / 216. Yes! That's exactly the probability given. So, n=4 is correct. Therefore, the answer is 4.

But let me check again my derivation. Because when I simplified the original equation, I had:

C(n, 2)*(5/6)^2*(1/6)^{n - 2} = 25/216, which led me to n=4. So that seems correct.

Alternatively, maybe there's another value? Let me check n=5, just to make sure. If n=5, what's the probability?

C(5, 2) = 10. Probability: 10*(5/6)^2*(1/6)^{3} = 10*(25/36)*(1/216). Wait, 25/36 * 1/216 is 25/(36*216). Then multiply by 10: 250/(36*216). Let me compute 36*216: 36*200=7200, 36*16=576, total 7200 + 576 = 7776. Therefore, probability is 250 / 7776. Let me divide numerator and denominator by 2: 125 / 3888. Which is approximately 0.03215... whereas 25/216 is approximately 0.1157. So, definitely different. So, n=5 is not the answer.

Similarly, n=3: C(3,2)=3. Probability: 3*(5/6)^2*(1/6)^{1} = 3*(25/36)*(1/6) = 75 / 216 = 25 / 72 ≈ 0.347, which is bigger than 25/216. So, yes, n=4 is the correct answer.

Wait, but just to make sure, let me check n=4 again. As above, 6*(25/36)*(1/36). Wait, no. Wait, (5/6)^2 is 25/36, (1/6)^{n - 2} is (1/6)^{2} which is 1/36. Multiply those together: 25/36 * 1/36 = 25/1296. Multiply by C(4, 2)=6: 6*25/1296=150/1296. Simplify: divide numerator and denominator by 6: 25/216. Exactly. So that's correct.

Therefore, n=4. So I think that's the answer. Let me check if there's a possibility of another value. For example, maybe n= something else?

Suppose n=6. Let's compute the probability. C(6,2)=15. Probability: 15*(5/6)^2*(1/6)^4. Let's compute:

(5/6)^2 = 25/36, (1/6)^4 = 1/1296. Multiply them: 25 / (36*1296). Multiply by 15: 375 / (36*1296). Let me compute denominator: 36*1296= 36*1296. 1296*36: 1296*30=38880, 1296*6=7776, total 38880 + 7776 = 46656. So 375 / 46656. Simplify: divide numerator and denominator by 3: 125 / 15552 ≈ 0.008. Which is way smaller than 25/216. So, nope. So n=4 is the only one.

Wait, but let me check n= something like 0? No. Negative? No. So n=4 is the only integer solution.

Therefore, I think the answer is 4. But let me just check if my algebra steps were correct. Starting from the equation:

Original equation:

C(n, 2) * (5/6)^2 * (1/6)^{n - 2} = 25/216

Then dividing both sides by (25/36):

C(n, 2) * (1/6)^{n - 2} = (25/216) / (25/36) = 1/6. Correct.

Then, substituting C(n,2):

[n(n -1)/2] * (1/6)^{n -2} = 1/6. Multiply both sides by 2:

n(n -1) * (1/6)^{n -2} = 1/3. Then, multiply both sides by 6^{n -2}:

n(n -1) = (1/3) * 6^{n -2}

Yes, then 6^{n -2} = 3n(n -1). Then, for n=4, 6^{2}=3*4*3 => 36=36. Correct. So all steps are correct.

Therefore, I think n=4 is indeed the correct answer.

**Final Answer**
The value of $ n $ is \boxed{4}.

To solve the problem, we are given that when $ n $ fair 6-sided dice are rolled simultaneously, the probability that **exactly two** of them show a number **other than 1** is:

$$
\frac{25}{216}
$$

We are to determine the value of $ n $.

---

### Step 1: Model the Situation

Each die roll has two possible outcomes:

- A success: the die shows a number **not equal to 1** (i.e., it shows 2, 3, 4, 5, or 6): probability $ \frac{5}{6} $
- A failure: the die shows a 1: probability $ \frac{1}{6} $

We want the probability that **exactly 2 out of $ n $** dice show a number other than 1. This is a **binomial probability problem**.

The probability mass function for a binomial distribution is:

$$
P(k) = \binom{n}{k} p^k (1 - p)^{n - k}
$$

Here:

- $ k = 2 $
- $ p = \frac{5}{6} $
- $ n $ is unknown

So the desired probability is:

$$
P(2) = \binom{n}{2} \left(\frac{5}{6}\right)^2 \left(\frac{1}{6}\right)^{n - 2}
$$

This is given to be:

$$
\frac{25}{216}
$$

---

### Step 2: Set Up the Equation

$$
\binom{n}{2} \left(\frac{25}{36}\right) \left(\frac{1}{6}\right)^{n - 2} = \frac{25}{216}
$$

Divide both sides by $ \frac{25}{36} $:

$$
\binom{n}{2} \left(\frac{1}{6}\right)^{n - 2} = \frac{1}{6}
$$

Substitute $ \binom{n}{2} = \frac{n(n - 1)}{2} $:

$$
\frac{n(n - 1)}{2} \cdot \left(\frac{1}{6}\right)^{n - 2} = \frac{1}{6}
$$

Multiply both sides by 2:

$$
n(n - 1) \cdot \left(\frac{1}{6}\right)^{n - 2} = \frac{1}{3}
$$

Multiply both sides by $ 6^{n - 2} $:

$$
n(n - 1) = \frac{1}{3} \cdot 6^{n - 2}
$$

---

### Step 3: Try Small Integer Values for $ n $

Let’s test small integer values:

- **For $ n = 2 $:**

$$
2 \cdot 1 = 2, \quad 6^{0} = 1 \Rightarrow \frac{1}{3} \cdot 1 = \frac{1}{3} \ne 2
$$

- **For $ n = 3 $:**

$$
3 \cdot 2 = 6, \quad 6^{1} = 6 \Rightarrow \frac{1}{3} \cdot 6 = 2 \ne 6
$$

- **For $ n = 4 $:**

$$
4 \cdot 3 = 12, \quad 6^{2} = 36 \Rightarrow \frac{1}{3} \cdot 36 = 12
$$

**Perfect match!** So, $ n = 4 $ satisfies the equation.

---

### Step 4: Verify the Probability for $ n = 4 $

We double-check the original probability formula:

$$
P(2) = \binom{4}{2} \left(\frac{5}{6}\right)^2 \left(\frac{1}{6}\right)^2
$$

$$
= 6 \cdot \frac{25}{36} \cdot \frac{1}{36} = \frac{150}{1296} = \frac{25}{216}
$$

This matches exactly with the given probability, confirming that $ n = 4 $ is indeed the correct value.

---

### Final Answer

$$
\boxed{4}
$$

[Assistant]
Explanation:
Okay, so I need to find the number of dice, n, such that when you roll them all at the same time, the probability that exactly two of them show a number other than 1 is 25/216. Hmm, let me try to break this down.

First, let me recall some probability concepts. When dealing with dice, each die has 6 sides, numbered 1 through 6. The problem says "exactly two of them show a number other than 1." So, that means out of the n dice, exactly two are not 1, and the rest (n - 2) are 1s. Wait, is that right? Let me check.

Yes, exactly two show a number other than 1, so the other n - 2 must show 1. So, the probability of this happening is calculated using the binomial probability formula. The binomial formula is P(k) = C(n, k) * p^k * (1 - p)^(n - k), where p is the probability of success on a single trial. In this case, a "success" would be rolling a number other than 1. Let me confirm.

Each die has a probability of 5/6 of showing a number other than 1, right? Because there are 5 numbers that are not 1 (2 through 6). And the probability of rolling a 1 is 1/6. So, if we're looking for exactly two dice that are not 1, then k = 2. So, substituting into the formula:

P(2) = C(n, 2) * (5/6)^2 * (1/6)^(n - 2)

And this is given to be equal to 25/216. So, we need to solve for n in the equation:

C(n, 2) * (5/6)^2 * (1/6)^(n - 2) = 25/216

Okay, let me write that down more clearly:

C(n, 2) * (25/36) * (1/6)^{n - 2} = 25/216

First, maybe I can simplify this equation. Let me note that 25/36 is (5/6)^2, and 25/216 is 25 divided by 6^3. Let me see if I can cancel some terms.

Divide both sides of the equation by 25/36:

C(n, 2) * (1/6)^{n - 2} = (25/216) / (25/36) = (25/216) * (36/25) = 36/216 = 1/6

So, simplifying, we have:

C(n, 2) * (1/6)^{n - 2} = 1/6

Hmm, so let me write that equation again:

C(n, 2) * (1/6)^{n - 2} = 1/6

Let me think about how to solve this equation for n. Since n is an integer greater than or equal to 2 (because we need exactly two dice to show non-1 numbers), we can try plugging in values of n and see which one satisfies the equation. Alternatively, maybe I can manipulate the equation algebraically.

First, let me recall that C(n, 2) is equal to n(n - 1)/2. So substituting that in:

[n(n - 1)/2] * (1/6)^{n - 2} = 1/6

Multiply both sides by 2:

n(n - 1) * (1/6)^{n - 2} = 2/6 = 1/3

So:

n(n - 1) * (1/6)^{n - 2} = 1/3

Hmm. Let me rewrite this equation as:

n(n - 1) / 6^{n - 2} = 1/3

Multiply both sides by 6^{n - 2}:

n(n - 1) = (1/3) * 6^{n - 2}

Alternatively:

n(n - 1) = 6^{n - 2} / 3

Let me note that 6^{n - 2} / 3 is equal to 6^{n - 2} * 3^{-1} = 6^{n - 2} / 3. Alternatively, since 6^{n - 2} = 6^{n} / 6^2 = 6^n / 36, so substituting back:

n(n - 1) = (6^n / 36) / 3 = 6^n / 108

Therefore:

n(n - 1) = 6^n / 108

Multiply both sides by 108:

108n(n - 1) = 6^n

Hmm. So, 6^n = 108n(n - 1). Let me note that 108 is 6^2 * 3, since 6^2 is 36 and 36*3 is 108. So, 6^n = 6^2 * 3 * n(n - 1). Therefore:

6^{n - 2} = 3n(n - 1)

So, 6^{n - 2} = 3n(n - 1). Hmm. Maybe this equation is easier to work with. Let me write that:

6^{n - 2} = 3n(n - 1)

Alternatively, divide both sides by 3:

2 * 6^{n - 2} = n(n - 1)

Wait, 6^{n - 2} divided by 3 is 6^{n - 2}/3 = 2^{n - 2} * 3^{n - 2}/3 = 2^{n - 2} * 3^{n - 3}. Maybe that's not helpful. Alternatively, let me try plugging in small integer values for n and see if I can find a solution.

Let me start with n=2. Let me check:

For n=2:

Left side: 6^{2 - 2} = 6^0 = 1

Right side: 3*2*1 = 6. 1 vs 6: Not equal.

n=3:

Left side: 6^{3 - 2} = 6^1 = 6

Right side: 3*3*2 = 18. 6 vs 18: Not equal.

n=4:

Left side: 6^{4 - 2} = 6^2 = 36

Right side: 3*4*3 = 36. Oh! 36 = 36. So, n=4 is a solution.

Wait, so n=4 works? Let me check if there are other solutions. Let me check n=5.

Left side: 6^{5 - 2} = 6^3 = 216

Right side: 3*5*4 = 60. 216 vs 60: Not equal.

n=6:

Left side: 6^{6 - 2} = 6^4 = 1296

Right side: 3*6*5 = 90. 1296 vs 90: Nope.

What about n=1? Wait, n must be at least 2 because we need exactly two dice to show non-1 numbers. If n=1, you can't have exactly two dice. So n=2 is the minimum. So n=4 is a solution. Let me check n=0? Not relevant. Negative numbers? No.

But let me check if there are other solutions. For example, n=5 gives left side 216 vs right side 60. So left side increases exponentially, while the right side is quadratic. So, for n=4, they are equal. For n>4, left side grows much faster. For n=3, left side is 6 vs right side 18. So for n=3, right side is bigger. So, n=4 is the only integer where they are equal?

Wait, is there another value? Let me check n=5 again. 6^3=216, right side 60. So left is bigger. What about n=4? 36 vs 36. For n=3, 6 vs 18. For n=2, 1 vs 6. So for n=4, it works, and for n>4, left side grows faster. So, only n=4?

But let me check if maybe there's another value where maybe the equation holds. Let me check n= something else? Like n= something smaller? But n has to be at least 2. For n=2, left side 1 vs 6. No. For n=3, 6 vs 18. No. So only n=4? So that suggests that n=4 is the answer. Let me check my steps again to make sure I didn't make a mistake.

Original probability was given as 25/216. Let me check if for n=4, the probability is indeed 25/216. Let me compute it again.

For n=4, probability that exactly two show numbers other than 1. So, number of ways: C(4, 2). Then, probability for each such case: (5/6)^2*(1/6)^{4-2} = (5/6)^2*(1/6)^2. Then, multiply by C(4, 2).

So, C(4, 2) is 6. Then 6*(25/36)*(1/36) ?

Wait, no. Wait: (5/6)^2 is 25/36, and (1/6)^{4-2} is (1/6)^2 = 1/36. Then multiply them together: 25/36 * 1/36 = 25 / 1296. Multiply by 6: 6 * 25 / 1296 = 150 / 1296. Simplify this fraction: divide numerator and denominator by 6: 25 / 216. Yes! That's exactly the probability given. So, n=4 is correct. Therefore, the answer is 4.

But let me check again my derivation. Because when I simplified the original equation, I had:

C(n, 2)*(5/6)^2*(1/6)^{n - 2} = 25/216, which led me to n=4. So that seems correct.

Alternatively, maybe there's another value? Let me check n=5, just to make sure. If n=5, what's the probability?

C(5, 2) = 10. Probability: 10*(5/6)^2*(1/6)^{3} = 10*(25/36)*(1/216). Wait, 25/36 * 1/216 is 25/(36*216). Then multiply by 10: 250/(36*216). Let me compute 36*216: 36*200=7200, 36*16=576, total 7200 + 576 = 7776. Therefore, probability is 250 / 7776. Let me divide numerator and denominator by 2: 125 / 3888. Which is approximately 0.03215... whereas 25/216 is approximately 0.1157. So, definitely different. So, n=5 is not the answer.

Similarly, n=3: C(3,2)=3. Probability: 3*(5/6)^2*(1/6)^{1} = 3*(25/36)*(1/6) = 75 / 216 = 25 / 72 ≈ 0.347, which is bigger than 25/216. So, yes, n=4 is the correct answer.

Wait, but just to make sure, let me check n=4 again. As above, 6*(25/36)*(1/36). Wait, no. Wait, (5/6)^2 is 25/36, (1/6)^{n - 2} is (1/6)^{2} which is 1/36. Multiply those together: 25/36 * 1/36 = 25/1296. Multiply by C(4, 2)=6: 6*25/1296=150/1296. Simplify: divide numerator and denominator by 6: 25/216. Exactly. So that's correct.

Therefore, n=4. So I think that's the answer. Let me check if there's a possibility of another value. For example, maybe n= something else?

Suppose n=6. Let's compute the probability. C(6,2)=15. Probability: 15*(5/6)^2*(1/6)^4. Let's compute:

(5/6)^2 = 25/36, (1/6)^4 = 1/1296. Multiply them: 25 / (36*1296). Multiply by 15: 375 / (36*1296). Let me compute denominator: 36*1296= 36*1296. 1296*36: 1296*30=38880, 1296*6=7776, total 38880 + 7776 = 46656. So 375 / 46656. Simplify: divide numerator and denominator by 3: 125 / 15552 ≈ 0.008. Which is way smaller than 25/216. So, nope. So n=4 is the only one.

Wait, but let me check n= something like 0? No. Negative? No. So n=4 is the only integer solution.

Therefore, I think the answer is 4. But let me just check if my algebra steps were correct. Starting from the equation:

Original equation:

C(n, 2) * (5/6)^2 * (1/6)^{n - 2} = 25/216

Then dividing both sides by (25/36):

C(n, 2) * (1/6)^{n - 2} = (25/216) / (25/36) = 1/6. Correct.

Then, substituting C(n,2):

[n(n -1)/2] * (1/6)^{n -2} = 1/6. Multiply both sides by 2:

n(n -1) * (1/6)^{n -2} = 1/3. Then, multiply both sides by 6^{n -2}:

n(n -1) = (1/3) * 6^{n -2}

Yes, then 6^{n -2} = 3n(n -1). Then, for n=4, 6^{2}=3*4*3 => 36=36. Correct. So all steps are correct.

Therefore, I think n=4 is indeed the correct answer.

**Final Answer**
The value of $ n $ is \boxed{4}.

To solve the problem, we are given that when $ n $ fair 6-sided dice are rolled simultaneously, the probability that **exactly two** of them show a number **other than 1** is:

$$
\frac{25}{216}
$$

We are to determine the value of $ n $.

---

### Step 1: Model the Situation

Each die roll has two possible outcomes:

- A success: the die shows a number **not equal to 1** (i.e., it shows 2, 3, 4, 5, or 6): probability $ \frac{5}{6} $
- A failure: the die shows a 1: probability $ \frac{1}{6} $

We want the probability that **exactly 2 out of $ n $** dice show a number other than 1. This is a **binomial probability problem**.

The probability mass function for a binomial distribution is:

$$
P(k) = \binom{n}{k} p^k (1 - p)^{n - k}
$$

Here:

- $ k = 2 $
- $ p = \frac{5}{6} $
- $ n $ is unknown

So the desired probability is:

$$
P(2) = \binom{n}{2} \left(\frac{5}{6}\right)^2 \left(\frac{1}{6}\right)^{n - 2}
$$

This is given to be:

$$
\frac{25}{216}
$$

---

### Step 2: Set Up the Equation

$$
\binom{n}{2} \left(\frac{25}{36}\right) \left(\frac{1}{6}\right)^{n - 2} = \frac{25}{216}
$$

Divide both sides by $ \frac{25}{36} $:

$$
\binom{n}{2} \left(\frac{1}{6}\right)^{n - 2} = \frac{1}{6}
$$

Substitute $ \binom{n}{2} = \frac{n(n - 1)}{2} $:

$$
\frac{n(n - 1)}{2} \cdot \left(\frac{1}{6}\right)^{n - 2} = \frac{1}{6}
$$

Multiply both sides by 2:

$$
n(n - 1) \cdot \left(\frac{1}{6}\right)^{n - 2} = \frac{1}{3}
$$

Multiply both sides by $ 6^{n - 2} $:

$$
n(n - 1) = \frac{1}{3} \cdot 6^{n - 2}
$$

---

### Step 3: Try Small Integer Values for $ n $

Let’s test small integer values:

- **For $ n = 2 $:**

$$
2 \cdot 1 = 2, \quad 6^{0} = 1 \Rightarrow \frac{1}{3} \cdot 1 = \frac{1}{3} \ne 2
$$

- **For $ n = 3 $:**

$$
3 \cdot 2 = 6, \quad 6^{1} = 6 \Rightarrow \frac{1}{3} \cdot 6 = 2 \ne 6
$$

- **For $ n = 4 $:**

$$
4 \cdot 3 = 12, \quad 6^{2} = 36 \Rightarrow \frac{1}{3} \cdot 36 = 12
$$

**Perfect match!** So, $ n = 4 $ satisfies the equation.

---

### Step 4: Verify the Probability for $ n = 4 $

We double-check the original probability formula:

$$
P(2) = \binom{4}{2} \left(\frac{5}{6}\right)^2 \left(\frac{1}{6}\right)^2
$$

$$
= 6 \cdot \frac{25}{36} \cdot \frac{1}{36} = \frac{150}{1296} = \frac{25}{216}
$$

This matches exactly with the given probability, confirming that $ n = 4 $ is indeed the correct value.

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### Final Answer

$$
\boxed{4}
$$

Solution: For $n$ dice, there are $\binom{n}{2}=\frac{n(n-1)}{2}$ ways to choose two of them. For each of these ways, there is a $\left( \frac{5}{6} \right)^2 \left( \frac{1}{6} \right)^{n-2}$ chance that all but the chosen two dice will roll a 1. Therefore, we need to find the value of $n$ for which $\frac{25n(n-1)}{2 \cdot 6^n}=\frac{25}{216}$, or $108n(n-1)=6^n$. Plugging in values for $n$, we can see that $n=\boxed{4}$ works and no value of $n$ less than 4 works. Now we just need to prove that no values of $n$ greater than 4 work. Note that if $n \geq 5$, then $n < 3^{n-3}$ and $n-1 < 2\cdot 2^{n-3}$. We can multiply these inequalities to get that when $n \geq 5$, we have $n(n-1) < 2\cdot 6^{n-3}$, or $108n(n-1)<6^n$.