TITLE: An integral inequality with inverse
QUESTION [8 upvotes]: 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.

REPLY [4 votes]: The inequality cannot hold since a non-decreasing concave function $f$ satisfing $f(0)=0$ and $f(1)=1$ can be almost everywhere arbitrarily close to one, with its inverse function being almost everywhere arbitrarily close to zero on $[0,1]$. Hence the given integral can be less than $\varepsilon$ for any $\varepsilon>0$. 
Examples of such a phenomenon are given by
$$ f(x) = \frac{(n+1)x}{n+x}\quad\mbox{for }n\geq 7, $$
$$ f(x) = x^r\quad\mbox{for }r\in(0,1/6),$$
$$ f(x) = x+\min\left(nx,\frac{n}{n+1}(1-x)\right)\quad\mbox{for }n\geq 4.$$