Proving the expected value of the square root of X is less than the square root of the expected value of X How do I show that $E(\sqrt{X}) \leq \sqrt{E(X)}$ for a positive random variable $X$?
I may be intended to use the Cauchy-Schwarz Inequality, $[E(XY)]^2 \leq E(X^2)E(Y^2)$, but I'm not sure how.

$\sqrt{x}$, $x\geq 0$ is a concave function so Jensen's inequality gives us the result without further effort.
