TITLE: How to find: $~\min\limits_{f\in E}(\int_0^1f(x) \,dx)$
QUESTION [10 upvotes]: I came across to the following  problem:

Let $E$ be the set of all continuous function $f:[0,1]\to \mathbb{R}$ such that $$f(x)+f(y)\ge |x-y|\qquad\forall\,x,y\in [0,1]$$
Then find $$\min_{f\in E}\left(\int_0^1f(x) dx\right)$$

My attempt:
I took the double integral on both side which yields $$ 2\int_0^1f(x)dx =\int_0^1\int_0^1f(x) +f(y)dydx \ge \int_0^1\int_0^1|x-y|dxdy =\frac{1}{3} $$
Thus, 
$$~\min\limits_{f\in E}(\int_0^1f(x) \,dx) \ge \frac{1}{6}$$
Unfortunately I don't know How to get the minimizer. Please give help me with a hint or an answer.
Minimize $\min_{f\in E}\left(\int_0^1f(x) dx\right)$

REPLY [22 votes]: Setting $y=1-x$, $f(x)+f(1-x)\geq |2x-1|$, and since $\int_0^1 f(1-x) dx =\int_0^1 f(x) dx$, integrate to get $$2\int_0^1 f(x)dx\geq \int_0^1 |2x-1| dx = \frac 12$$
hence $\int_0^1 f(x) dx \geq \frac 14$.
This bound is attained for $f:x\mapsto |x-\frac 12|$. The triangle inequality trivially yields $f(x)+f(y)\ge |x-y|$ and $\int_0^1 f(x)dx= \frac 14$