If $X_1$ and $X_2$ are random variables, then $E(X_1^2)E(X_2^2) \geq [E(X_1X_2)]^2$ by Schwarz's inequality. Use this fact to show that $-1 \leq \rho_{12} \leq 1$.
I know that $\rho_{12} = \frac{(E(X_1X_2) - E(X_1)E(X_2))}{ \sqrt{(E(X_1^2) - E(X_1)^2)(E(X_2^2) -E(X_2)^2)}}$
But I am stuck from this point on. I believe I need to show that the denominator is always smaller than the denominator, which would show that it would never be greater than 1. Squaring the whole thing gives me some $E(X_1^2)E(X_2^2)$ and $[E(X_1X_2)]^2$ terms in the numerator, which were mentioned in the inequality, but the rest gets really ugly.
Any suggestions on where to go from here?
Recall that the definition of the variance of $X$ is $\text{Var}(X)=E((X-\mu_X)^2)$. Your version $E(X^2)-(\mu_X)^2$ happens to be often more computationally convenient. It is not hard to show they are equal.
Similarly, the covariance of $X$ and $Y$ is defined as $E((X-\mu_X)(Y-\mu_Y))$. Your version is equivalent, and the equivalence is not hard to prove.
If you use the definitions of variance and covariance, the result follows directly from Cauchy-Schwarz.
If you want to use the formulas for variance and covariance that you quoted, a first step is to prove the equivalence to the basic definitions. Or else you can show that variance and covariance are preserved by shifts, and shift so as to make the means $0$.