MHB Prove: Integral Inequality for Convex Function

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The discussion centers on proving the integral inequality for a convex function defined on the interval [1, 13]. It states that for a convex and integrable function f, the sum of the integrals from 1 to 3 and from 11 to 13 must be greater than or equal to the integral from 5 to 9. Participants explore the properties of convex functions and their implications for integrals over specified intervals. The proof relies on the definition of convexity and properties of integrals. Ultimately, the inequality highlights the behavior of convex functions in relation to their integrals over different intervals.
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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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