Prove that random variables satisfy the inequality $E(XY)^2 \le E(X^2)E(Y^2)$? Given Random variables $X$ and $Y$ is it true always that;
$$E(XY)^2 \le  E(X^2)E(Y^2)$$
Is it easy to prove?
 A: 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
A: 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.
A: This is known as the Cauchy Schwarz inequality for Random Variables.
