MHB Prove: Integral Inequality for Convex Function

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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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