For $f$ convex $\int_0^af+\int_0^bf+\int_0^cf+\int_0^{a+b+c}f\geq \int_0^{a+b}f+\int_0^{b+c}f+\int_0^{c+a}f$ The following problem was shortlisted for the mathematical olympiad in Romania a few years ago.

Let $f : [0,\infty) \to \mathbb{R}$ be a convex function and $a, b, c$ positive numbers, then
$$\int_0^af(x)dx+\int_0^bf(x)dx+\int_0^cf(x)dx+\int_0^{a+b+c}f(x)dx\geq \int_0^{a+b}f(x)dx+\int_0^{b+c}f(x)dx+\int_0^{c+a}f(x)dx$$

The only idea I had in order to put to use the convexity of $f$ was the well known inequality
$$\frac{f(a)+f(b)}{2}\geq \frac{1}{b-a}\int_a^bf(x)dx\geq f\left(\frac{a+b}{2}\right)$$
where $f$ is a convex function but after getting rid of the integrals I'm not even sure if that inequality always holds.
 A: For $a,b,c=0$, we have $0=0$ equality. Now, consider
$$G(a,b,c) = \int_0^a f+\int_0^b f+\int_0^c f+\int_0^{a+b+c}f-\int_0^{a+b}f-\int_0^{b+c}f-\int_0^{c+a}f$$
Differentiate it. By the fundamental theorem of calculus (We can apply it because a convex function is necessarily continuous except possibly at endpoints of the domain):
\begin{align*}\frac{\partial G}{\partial a} &= f(a)+f(a+b+c)-f(a+b)-f(a+c)\\
&\ge f(a)+f(a+b+c)-\left(\frac{c}{b+c}f(a)+\frac{b}{b+c}f(a+b+c)\right)\\
&\quad-\left(\frac{b}{b+c}f(a)+\frac{c}{b+c}f(a+b+c)\right)\\
\frac{\partial G}{\partial a} &\ge 0\end{align*}
The inequality there holds for nonnegative $b$ and $c$, using the characterization that convex functions lie below their secant lines. By symmetry, then, $\frac{\partial G}{\partial b}$ and $\frac{\partial G}{\partial c}$ are also nonnegative when $a,b,c\ge 0$. Together with the already established fact that $G(0,0,0)=0$, we get that $G(a,b,c)\ge 0$ for $a,b,c\ge 0$. If you want an explicit proof of that, we can apply the Mean Value theorem on the three segments $(0,0,0)$ to $(a,0,0)$, $(a,0,0)$ to $(a,b,0)$, and $(a,b,0)$ to $(a,b,c)$.
