For a non-negative random variable $X$ with distribution function $F$, show that $\int_0^{\infty} (1 − F(t))^2 dt \leq (E (\sqrt{X}))^2$?
I was using Holder's inequality to get $\sqrt{E(X^2)} \leq E(\sqrt{X})^2$ but can't proceed further to obtain the LHS. I know $E(X)=\int_0^{\infty} (1-F(t))dt$
Best Answer
Note that $\min\{x,y\}\leq \sqrt x\sqrt y$ for nonnegative $x,y$. If $X$ and $Y$ are IID, then LHS is $\mathbb E[\min\{X,Y\}]$ while the RHS is $\mathbb E[\sqrt X \sqrt Y]$.