$x\in [0,1]$, $f(0)=0$, $f(1)=1$,
$f$ is increasing and weakly concave
$$\int_0^1f(x)[f(x)-y]\,\mathrm dx=0$$
We want to show that $y\geq\dfrac{2}{3}.$
My (complicated) approach is to consider the limiting case first: $f(x)$ is linear
Then the integral becomes $\dfrac{1}{3}1^3-\dfrac{1}{2}1^2\cdot y'=0$
We have $y'=2/3$.
Then we may prove that $y\,(f \text{ is concave})>y'\,(f \text{ is linear})$.
Does this approach works? Do you have smarter ideas?
Solve for $y$:
$$y=\dfrac{\displaystyle\int_0^1f^2\,\mathrm dx}{\displaystyle\int_0^1f\,\mathrm dx}$$
Let $g(x)=x$ be the linear function that I used above.
It seems like if $f$ is concave, then $f(x)\geq g(x)$, and
$f^2/f\geq g^2/g$ for any $x$.
Can we conclude that $\dfrac{\displaystyle\int_0^1f^2\,\mathrm dx}{\displaystyle\int_0^1f\,\mathrm dx}\geq\dfrac{\displaystyle\int_0^1g^2\,\mathrm dx}{\displaystyle\int_0^1g\,\mathrm dx}$ ?
Title edited. Thanks for comments! It is derived from an old practice contest question.