Show that $\int_1^3f(x)dx+\int_{11}^{13}f(x)dx\ge\int_5^9f(x)dx$

Question

Let $f:[1,13]\to\mathbb{R}$ be a convex and integrable function. Show that $$\int_1^3f(x)dx+\int_{11}^{13}f(x)dx\ge\int_5^9f(x)dx$$

The solution comes pretty fast if we use the following inequality and some substitutions
(*) $f:I\to\mathbb{R}$, f convex on $I$. For every $a, b, c\in I$; $a\lt b\lt c$, we have:
$f(a-b+c)\le f(a)-f(b)+f(c)$

I was wondering if there's a way to solve this problem without using that inequality

quick solution with that inequality:
Let $b=a+4; c=a+10\implies f(a+6)+f(a+4)\le f(a)+f(a+10)$ We integrate this from 1 to 3
$\int_1^3f(a+6)+\int_1^3f(a+4)\le\int_1^3f(a)+\int_1^3f(a+10)$
We make the following substitution for every integral $x = a+6; x=a+4; x=a; x=a+10$ and that's all

Answer 1

Let $l$ be the straight line through the points $(5,f(5)), (9,f(9))$.

Note that $\int_1^3 l(x)dx + \int_{11}^{13} l(x)dx = \int_5^9 l(x)dx$.

Since $f$ is convex, $f(x) \le l(x)$ for $x \in [5,9]$ and $f(x) \ge l(x)$ otherwise.