Show that $E^2(XY) \leq E(X^2)E(Y^2)$.
We can assume that $X$ and $Y$ are jointly distributed random variables with finite variances.
I know cov$(X,Y)=E[(X-\mu_x)(Y-\mu_y)]$ and there's a definition that might be of use that states $V(X)=E(X^2)-E(X)^2$.
Any idea where to start to prove this?
In case @angryavian's insight is unintuitive:
Let $Z\in\{X,\,Y\}$ have mean $\mu_Z$ and standard deviation $\sigma_Z$ and let$$\rho:=\operatorname{Corr}(X,\,Y)\in[-1,\,1].$$Applying sign changes if necessary so means are non-negative,$$\begin{align}E(X^2)E(Y^2)-E(XY)^2&=(\mu_X^2+\sigma_X^2)(\mu_Y^2+\sigma_Y^2)-(\mu_X\mu_Y+\rho\sigma_X\sigma_Y)^2\\&=\sigma_X^2\mu_Y^2+\mu_X^2\sigma_Y^2-2\rho\mu_X\mu_Y\sigma_X\sigma_Y+(1-\rho^2)\sigma_X^2\sigma_Y^2\\&\ge\sigma_X^2\mu_Y^2+\mu_X^2\sigma_Y^2-2\mu_X\mu_Y\sigma_X\sigma_Y\\&=(\mu_X\sigma_Y-\mu_Y\sigma_X)^2.\end{align}$$