Let's analyze the problem and prove the given inequality step by step.

Problem Statement Recap:

We are given a function $f(x)$ that is monotonically increasing, strictly concave, and twice differentiable. The function is defined on the interval $[h, k]$, and we need to prove that:

$\frac{1}{b-a} \int_{a}^{b} f(x) \, dx > \frac{1}{k-h} \int_{h}^{k} f(x) \, dx,$

given that $a + b > h + k$, and $h < a < b < k$ are arbitrary constants.

Steps for Proof:

1. Mean Value Theorem for Integrals:

By the mean value theorem for integrals, there exist points $c_1 \in [a, b]$ and $c_2 \in [h, k]$ such that: $\frac{1}{b-a} \int_{a}^{b} f(x) \, dx = f(c_1),$ $\frac{1}{k-h} \int_{h}^{k} f(x) \, dx = f(c_2).$

Thus, the inequality we need to prove reduces to: $f(c_1) > f(c_2).$

2. Behavior of $f(x)$:

Since $f(x)$ is strictly concave, its first derivative $f'(x)$ is decreasing, and because $f(x)$ is monotonically increasing, $f'(x) > 0$ on $[h, k]$.

Given that $a + b > h + k$, it implies that the midpoint $\frac{a+b}{2} > \frac{h+k}{2}$.

Since $f(x)$ is strictly concave, the average value of the function over any subinterval closer to the larger values of $x$ will be greater. Specifically, since $c_1$ lies in the interval $[a, b]$ and $c_2$ lies in $[h, k]$ and $\frac{a+b}{2} > \frac{h+k}{2}$, we have $c_1 > c_2$.

3. Monotonicity and Strict Concavity:

Because $f(x)$ is monotonically increasing, $c_1 > c_2$ implies $f(c_1) > f(c_2)$.

This leads directly to: $\frac{1}{b-a} \int_{a}^{b} f(x) \, dx = f(c_1) > f(c_2) = \frac{1}{k-h} \int_{h}^{k} f(x) \, dx.$

Conclusion:

Hence, we have proven that: $\frac{1}{b-a} \int_{a}^{b} f(x) \, dx > \frac{1}{k-h} \int_{h}^{k} f(x) \, dx,$ under the given conditions.

Do you have any questions or need further details on this proof?

Here are 5 related questions to further explore this topic:

1. How would the proof change if $f(x)$ was convex instead of concave?
2. Can you find an example of a function $f(x)$ that satisfies the conditions and verify the inequality numerically?
3. What is the geometric interpretation of the inequality involving the average value of $f(x)$ over different intervals?
4. How does the result change if we assume $a + b = h + k$?
5. Can this result be extended to higher dimensions for functions of several variables?

Tip: Understanding the behavior of a function's first and second derivatives is crucial when dealing with inequalities involving concave or convex functions.