How to find: $~\min\limits_{f\in E}(\int_0^1f(x) \,dx)$
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 $\int_0^1 f(x)dx= \frac 14$