For an arbitrary, monotonically increasing, strictly concave, and twice di¤erentiable function
f(x) of a continuous variable x; dened for h  x  k; prove that
1
b 􀀀 a
Z b
x=a
f(x)dx >
1
k 􀀀 h
Z k
x=h
f(x)dx; if a + b > h + k;
where h < a < b < k are arbitrary constants.

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.

For an arbitrary, monotonically increasing, strictly concave, and twice differentiable function f(x) of a continuous variable x; defined for h ≤ x ≤ k; prove that (1/(b - a)) ∫[a, b] f(x) dx > (1/(k - h)) ∫[h, k] f(x) dx, if a + b > h + k; where h < a < b < k are arbitrary constants.

Explore the proof of an inequality involving a monotonically increasing and strictly concave function. This problem uses advanced calculus concepts such as the Mean Value Theorem and properties of concave functions. Suitable for undergraduate students studying advanced calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation