Expectation inequality

Question

How can I prove that $$ \mathbb{E} [X^2] \geq \mathbb{E} [|X|]^2 $$ This resembles the Cauchy-Schwarz inequality a lot but I'm unable to prove it with the usual method (i.e. when there are two random variables and you choose to set $f(\alpha)=(X-\alpha Y)^2$. Any help is greatly appreciated!

Answer 1

From Jensen's inequality we know that, whenever f is a convex function and X is a random variable we have: $$ E(f(X)) \geq f(E(X))$$.

In this context $f(X) = X^2$, which is a convex function and hence we are done.

Answer 2

For $a,b\in\mathbb R$ and $t\in(0,1)$, we have \begin{align} (ta^2+(1-t)b^2) - (ta+(1-t)b)^2 &= ta^2 + (1-t)b^2 -t^2a^2 -2t(1-t)ab -(1-t)^2b^2\\ &= t(1-t)a^2 -2t(1-t)ab + t(1-t)b^2\\ &= t(1-t)(a-b)^2\\ &>0, \end{align} so $x\mapsto x^2$ is a convex function. It follows from Jensen's inequality that $$\mathbb E[X^2]\geqslant \mathbb E[X]^2 $$ and $$\mathbb E[X^2]=\mathbb E[(-X)^2]\geqslant \mathbb E[-X]^2, $$ and so $$\mathbb E[X^2]\geqslant \mathbb E[|X|]^2. $$