Okay, let me try to figure out this problem step by step. So, we have a quadratic polynomial P(x) that satisfies two inequalities for all real numbers x: x² - 2x + 2 ≤ P(x) ≤ 2x² - 4x + 3. Additionally, we know that P(11) = 181, and we need to find P(16). Hmm, okay. Let's break this down.

First, since P(x) is a quadratic polynomial, it has the general form P(x) = ax² + bx + c, where a, b, and c are real coefficients. The problem gives us two quadratic inequalities that P(x) must satisfy for all real x. So, we need to find a quadratic that lies between these two quadratics for every x. Also, we have a specific value at x = 11, which is P(11) = 181. Our goal is to determine P(16).

Let me start by analyzing the given inequalities. The lower bound is x² - 2x + 2, and the upper bound is 2x² - 4x + 3. Let's denote the lower quadratic as L(x) = x² - 2x + 2 and the upper quadratic as U(x) = 2x² - 4x + 3. So, for all real x, we have L(x) ≤ P(x) ≤ U(x).

Since P(x) is a quadratic polynomial, and L(x) and U(x) are also quadratics, the difference between P(x) and L(x) must be non-negative for all x, and the difference between U(x) and P(x) must also be non-negative for all x. Let me define these differences:

Let’s set Q(x) = P(x) - L(x). Then Q(x) = (ax² + bx + c) - (x² - 2x + 2) = (a - 1)x² + (b + 2)x + (c - 2). Since Q(x) ≥ 0 for all x, this quadratic must be non-negative for all real numbers. Similarly, let’s define R(x) = U(x) - P(x) = (2x² - 4x + 3) - (ax² + bx + c) = (2 - a)x² + (-4 - b)x + (3 - c). Since R(x) ≥ 0 for all x, this quadratic must also be non-negative for all real numbers.

For a quadratic to be non-negative for all real x, two conditions must hold: the leading coefficient must be non-negative, and the discriminant must be less than or equal to zero. So, applying this to both Q(x) and R(x):

For Q(x):
1. Leading coefficient: (a - 1) ≥ 0 ⇒ a ≥ 1.
2. Discriminant: [ (b + 2) ]² - 4*(a - 1)*(c - 2) ≤ 0.

For R(x):
1. Leading coefficient: (2 - a) ≥ 0 ⇒ a ≤ 2.
2. Discriminant: [ (-4 - b) ]² - 4*(2 - a)*(3 - c) ≤ 0.

So combining these, we have 1 ≤ a ≤ 2.

Now, since P(x) is sandwiched between L(x) and U(x), and both differences Q(x) and R(x) are non-negative quadratics, perhaps there's a way to express P(x) as a linear combination of L(x) and U(x)? Let me think. If P(x) is between L(x) and U(x) for all x, maybe it can be expressed as P(x) = t(x) * L(x) + (1 - t(x)) * U(x), where t(x) is a function between 0 and 1. But since P(x) is a quadratic, and L(x) and U(x) are quadratics, t(x) would have to be a constant for the combination to remain quadratic. Wait, if t is a constant, then P(x) would be a convex combination of L(x) and U(x). But maybe that's not necessarily the case here. Let me check.

If P(x) is a convex combination of L(x) and U(x), then it would be P(x) = k * L(x) + (1 - k) * U(x), where 0 ≤ k ≤ 1. Then, expanding this:

P(x) = k(x² - 2x + 2) + (1 - k)(2x² - 4x + 3)
= kx² - 2kx + 2k + 2(1 - k)x² - 4(1 - k)x + 3(1 - k)
= [k + 2(1 - k)]x² + [-2k - 4(1 - k)]x + [2k + 3(1 - k)]
= [2 - k]x² + [-2k -4 +4k]x + [2k +3 -3k]
= [2 - k]x² + [2k -4]x + [3 - k]

But then, since P(x) must also satisfy the given inequalities, maybe this is a possible approach. Let's see if this works. Then, we can check the discriminants for Q(x) and R(x) in this case.

Alternatively, maybe there's another approach. Let's think about the points where L(x) and U(x) are equal. Let's set L(x) = U(x):

x² - 2x + 2 = 2x² - 4x + 3

Subtract L(x) from both sides:

0 = x² - 2x + 1

Which factors as (x - 1)^2 = 0. So, they are equal only at x = 1. Therefore, the two quadratics touch at x = 1, and between them, P(x) must lie between L(x) and U(x) everywhere else. Since L(x) and U(x) intersect only at x = 1, and for all other x, one is above the other. Let's check which quadratic is above which. Let's pick a value, say x = 0:

L(0) = 0 - 0 + 2 = 2

U(0) = 0 - 0 + 3 = 3

So at x = 0, L(x) = 2, U(x) = 3. So L(x) < U(x) here. Let's check x = 2:

L(2) = 4 - 4 + 2 = 2

U(2) = 8 - 8 + 3 = 3

Same as above. So perhaps for x ≠ 1, L(x) < U(x). Let's check x approaching infinity. For large x, the leading term of L(x) is x² and U(x) is 2x², so U(x) is above L(x) for large x. Similarly, for x approaching negative infinity, the leading terms are x² and 2x², so again U(x) is above. Therefore, L(x) ≤ U(x) for all x, and they are equal only at x = 1. So the region between them is the area between these two curves, and P(x) must lie within that region for all x.

Therefore, P(x) is a quadratic that lies between L(x) and U(x) for all x, touching them at x = 1. So, perhaps P(x) is equal to L(x) plus some non-negative quadratic, or U(x) minus some non-negative quadratic. But since both Q(x) and R(x) are non-negative, and since Q(x) = P(x) - L(x) and R(x) = U(x) - P(x), both of which are non-negative quadratics.

Given that both Q(x) and R(x) are non-negative quadratics, their sum Q(x) + R(x) = U(x) - L(x). Let's compute U(x) - L(x):

U(x) - L(x) = (2x² - 4x + 3) - (x² - 2x + 2) = x² - 2x + 1 = (x - 1)^2. So, Q(x) + R(x) = (x - 1)^2. Since Q(x) and R(x) are both non-negative, they must each be proportional to (x - 1)^2. Wait, but (x - 1)^2 is a perfect square, so maybe Q(x) = k(x - 1)^2 and R(x) = (1 - k)(x - 1)^2 for some k between 0 and 1. Then Q(x) + R(x) = (x - 1)^2. That makes sense. Let's check this.

If Q(x) = k(x - 1)^2, then P(x) = L(x) + Q(x) = x² - 2x + 2 + k(x² - 2x + 1) = (1 + k)x² + (-2 - 2k)x + (2 + k). Similarly, R(x) = (1 - k)(x - 1)^2. Then, U(x) = P(x) + R(x) = [x² - 2x + 2] + k(x - 1)^2 + (1 - k)(x - 1)^2 = x² - 2x + 2 + (x - 1)^2 = 2x² - 4x + 3, which matches. So this seems promising.

Therefore, if we set Q(x) = k(x - 1)^2, then P(x) = L(x) + k(x - 1)^2. Similarly, since R(x) = (1 - k)(x - 1)^2, which is non-negative as long as 0 ≤ k ≤ 1. So this parameterization works.

Therefore, P(x) can be written as:

P(x) = L(x) + k(x - 1)^2 = (x² - 2x + 2) + k(x² - 2x + 1) = (1 + k)x² + (-2 - 2k)x + (2 + k)

Given that, we can express P(x) in terms of the parameter k. Then, since we have P(11) = 181, we can plug x = 11 into this expression to solve for k. Once we have k, we can write the entire polynomial P(x), and then compute P(16).

Let me proceed with this plan.

First, express P(11):

P(11) = (1 + k)(11)^2 + (-2 - 2k)(11) + (2 + k) = 181

Compute each term:

(1 + k)(121) = 121 + 121k

(-2 - 2k)(11) = -22 - 22k

(2 + k) = 2 + k

So sum all these:

121 + 121k -22 -22k + 2 + k = 181

Combine like terms:

121 -22 + 2 = 101

121k -22k + k = (121 -22 +1)k = 100k

So overall:

101 + 100k = 181

Subtract 101:

100k = 80

Divide by 100:

k = 0.8

So k = 4/5. Therefore, the polynomial is:

P(x) = (1 + 4/5)x² + (-2 - 2*(4/5))x + (2 + 4/5)

Simplify each term:

1 + 4/5 = 9/5

-2 - 8/5 = (-10/5 - 8/5) = -18/5

2 + 4/5 = 14/5

Therefore, P(x) = (9/5)x² - (18/5)x + 14/5

Let me verify this with P(11):

(9/5)(121) - (18/5)(11) + 14/5 = (1089/5) - (198/5) + 14/5 = (1089 - 198 + 14)/5 = (905)/5 = 181. Yes, that's correct.

Also, check if P(x) lies between L(x) and U(x) for some x. For example, at x = 1, P(1) should equal both L(1) and U(1):

L(1) = 1 - 2 + 2 = 1

U(1) = 2 - 4 + 3 = 1

P(1) = (9/5)(1) - (18/5)(1) + 14/5 = (9 - 18 + 14)/5 = 5/5 = 1. So that's correct.

Another test point: x = 0

L(0) = 2

U(0) = 3

P(0) = 14/5 = 2.8, which is between 2 and 3.

At x = 2:

L(2) = 4 - 4 + 2 = 2

U(2) = 8 - 8 + 3 = 3

P(2) = (9/5)(4) - (18/5)(2) + 14/5 = 36/5 - 36/5 + 14/5 = 14/5 = 2.8. Which is between 2 and 3.

Seems okay. So this seems to be the correct polynomial.

Therefore, P(16) can be calculated as:

P(16) = (9/5)(16)^2 - (18/5)(16) + 14/5

Compute each term:

(9/5)(256) = (2304)/5 = 460.8

(18/5)(16) = (288)/5 = 57.6

14/5 = 2.8

So:

460.8 - 57.6 + 2.8 = (460.8 - 57.6) + 2.8 = 403.2 + 2.8 = 406

Therefore, P(16) = 406. Let me check the arithmetic again to be sure.

First, 16 squared is 256. Multiply by 9/5: 256 * 9 = 2304; 2304 / 5 = 460.8.

16 multiplied by 18/5: 16*18 = 288; 288 /5 = 57.6.

14/5 is 2.8.

So 460.8 - 57.6 = 403.2; 403.2 + 2.8 = 406. Yes, that's correct.

Therefore, the final answer is 406. So, \boxed{406}.

**Final Answer**
\boxed{406}
Therefore, the value of $P(16)$ is $\boxed{406}$.
</think>