Mathematical Reflections 2016, Issue 1 - Problem U366

Problem:
If $f:[0,1] \to \mathbb{R}$ is a convex and integrable function with $f(0)=0$, prove that
$$\int_{0}^1 f(x) \ dx \geq 4\int_{0}^{\frac{1}{2}} f(x) \ dx.$$

Proposed by Florin Stanescu, Gaesti, Romania

Solution:
Since $f$ is a convex function, we have $$\renewcommand{\arraystretch}{2} \begin{array}{lll} \displaystyle \int_{0}^1 f(x) \ dx= \displaystyle \dfrac{1}{2}\left[\int_{0}^1 f(x) \ dx+\int_{0}^1 f(1-x) \ dx \right]&=&\displaystyle \int_{0}^1 \dfrac{f(x)+f(1-x)}{2} \ dx \\ &\geq& \displaystyle \int_{0}^1 f\left(\dfrac{x+(1-x)}{2}\right) \ dx\\&=&f\left(\dfrac{1}{2}\right). \end{array}$$
On the other hand, since $f$ is convex and $f(0)=0$, we have $$\renewcommand{\arraystretch}{2} \begin{array}{lll} \displaystyle f\left(\dfrac{1}{2}\right)=2\cdot\dfrac{f(0)+f\left(\frac{1}{2}\right)}{2}&=& \displaystyle 2\int_{0}^1 \left[(1-x)f(0)+xf\left(\dfrac{1}{2}\right)\right] \ dx \\ & \geq & \displaystyle 2\int_{0}^1 f\left((1-x)\cdot0+x\cdot\dfrac{1}{2}\right) \ dx \\&=& \displaystyle 4\int_{0}^{\frac{1}{2}} f(x) \ dx, \end{array}$$ and the conclusion follows.