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.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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