Prove: Integral Inequality for Convex Function

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SUMMARY

The discussion centers on proving the integral inequality for a convex function defined on the interval [1, 13]. Specifically, it establishes that for a convex and integrable function \( f \), the inequality \( \int_1^3 f(x)dx + \int_{11}^{13} f(x)dx \ge \int_5^9 f(x)dx \) holds true. This conclusion leverages the properties of convex functions and their behavior over specified intervals. The proof utilizes the definition of convexity and the properties of integrals to demonstrate the validity of the inequality.

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Let $f:[1,\,13]\rightarrow R$ be a convex and integrable function. Prove that $\displaystyle \int_1^3 f(x)dx+\int_{11}^{13} f(x)dx\ge \int_5^9 f(x)dx$,
 
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If $a<b<c$, we then have $f(a-b+c)+f(b)\le f(a)+f(c)$.

Let $c=a+10$ and $b=a+4$. Then we have

$f(a+6)+f(a+4)\le f(a)+f(a+10)$

If we integrate both sides for $a\in [1,\,3]$, and notice that

$\displaystyle \int_1^3 f(a+6)da=\int_{7}^{9} f(x)dx,\\ \displaystyle\int_1^3 f(a+4)da=\int_{5}^{7} f(x)dx$
and
$\displaystyle \int_1^3 f(a+10)da=\int_{11}^{13} f(x)dx$

the result then follows.
 

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