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.

Consider the function $f(x)=\sqrt{x}$.
As this is clearly a concave function, Jensen's inequality states 
$$
\mathbb{E}\Big[f(X)\Big]\leq f \Big(\mathbb{E}[X] \Big)
$$
and there you go
