We know that $$E[x^2] > E[x]^2$$ by Cauchy-Schwarz. Further we know that $$E[x^2]E[\sqrt{x}] > E[x]E[x^{1.5}]$$ when $x$ is restricted to positive reals by Callebaut's Inequality or Hölder's inequality. Can we prove that for all concave and positive functions $f$ such that $f(0) = 0$ we have $$E[x^2]E[f(x)] > E[x]E[xf(x)]$$
where $x$ is only supported on $(0, 1)$.
The inequalities are large.
I denote by $X$ the random variable, and by $Y$ an independent copy of $X$. I only assume that $X$ is non-negative and in $L^2$, so all expectations are finite.
We note that \begin{eqnarray*} 2E[X^2]E[f(X)] - 2E[X]E[Xf(X)] &=& E[X^2f(Y)] - E[XYf(Y)] \\ & & + E[Y^2f(X)] - E[YXf(X)] \\ &=& E[(X-Y)(Xf(Y)-Yf(X))]. \end{eqnarray*}
Because of the concavity of $f$ and the assumption $f(0)=0$, the random variable $(X-Y)(Xf(Y)-Yf(X)$ is null on the event $[XY=0]$ and non-negative on the event $[XY>0]$ since $f(Y)/Y-f(X)/X$ has the same sign as $X-Y$.
The desired inequality follows.