An integral inequality with inverse Let $f:[0,1]\to [0,1]$ be a non-decreasing concave function, such that $f(0)=0,f(1)=1$. Prove or disprove that :
$$ \int_{0}^{1}(f(x)f^{-1}(x))^2\,\mathrm{d}x\ge \frac{1}{12}$$
A friend posed this to me. He hopes to have solved it, but he is not sure. Can someone help? Thanks. 

I am not so sure this works. Take $f(x)=nx$ for $x \in [0;\frac{1}{n+1}]$ and $f(x)=\frac{n}{n+1} + (x-\frac{1}{n+1})\frac{1-\frac{n}{n+1}}{1-\frac{1}{n+1}} = \frac{n}{n+1} + \frac{1}{n}(x-\frac{1}{n+1})$ on $]\frac{1}{n+1},1]$.
Then $f^{-1}(y)= \frac{y}{n}$ if $y \in [0,\frac{n}{n+1}]$ and $f^{-1}(y)= n(y-\frac{n}{n+1})+\frac{1}{n+1}$ if $y \in [\frac{n}{n+1},1]$
Then, noticing that$\forall x$  $f(x)\leq 1$ and $f^{-1}(x)\leq 1$, we can rewrite:
$\int_0^1 (f f^{-1} )^2\leq \int_0^1 f^{-1}(x )^2dx \leq \int_0^{\frac{n}{n+1}}\frac{x^2}{n^2}dx + \int_{\frac{n}{n+1}}^1 1dx \leq \frac{1}{n ^2} + \frac{1}{n+1}$ which converges towards 0.
