Inequality with Expectations

I'm going to assume you mean $$(E(XY))^2 \le E(X^2)E(Y^2).$$ One way to prove this is to realize it's a special case of the Cauchy–Schwarz inequality.

Here's another. Let \begin{align} f(t) = {} & E((tX+Y)^2) \\[8pt] = {} & (E(X^2)) t^2 + 2(E(XY))t + E(Y^2) \\[8pt] = {} & at^2 + bt + c. \end{align} where $t$ is "constant", i.e. not random. Clearly $E((tX+Y)^2)\ge0$ for all real values of $t$. Now recall that for real $a,b,c$, the polynomial $at^2 + bt+c$ remains non-negative as $t$ changes if and only if $a\ge0$ and the discriminant $b^2-4ac\le0$. So $$ b^2-4ac = 4E(XY)^2 - 4E(X^2)E(Y^2). $$ So $$ 4(E(XY)^2 - E(X^2)E(Y^2))\le0. $$ Divide both sides by $4$ and there you have it.

The expectation of a product of random variables is an inner product, to which you can apply the Cauchy-Schwarz inequality and obtain exactly that inequality. Hence the answer is yes.

See http://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality#Probability_theory