Variance of Z for Z = X + Y, when X and Y correlated

So I'm trying to show that ${\rm Var}(Z) \le 2({\rm Var}(X)+{\rm Var}(Y))$ for $Z = X + Y$. This seems to be pretty easy to show given that $X$ and $Y$ are uncorrelated. But I'm running into trouble at this step: $${\rm Var}(Z) = {\rm Var}(X) + {\rm Var}(Y) + 2E[XY] - 2E[X]E[Y]$$ Normally you could say, $X$, $Y$ uncorrelated $\rightarrow E[XY] = E[X]E[Y]$, but when you cannot do this, I'm lost. Any tips?

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I don't want to write out the full answer because this looks a lot like homework or self-study (and if it is indeed homework or self-study, please add the homework or self-study tag).
Hint: the maximum value that $\operatorname{cov}(X,Y)$ can have is $\sqrt{\operatorname{var}(X)\cdot\operatorname{var}(Y)}$ (the minimum value is $-\sqrt{\operatorname{var}(X)\cdot\operatorname{var}(Y)}$). Use this together with $$\operatorname{var}(X\pm Y) = \operatorname{var}(X)+\operatorname{var}(Y) \pm 2\operatorname{cov}(X,Y)$$ to see if you can get anywhere with this exercise.
$\max \sigma_Z^2 = \sigma_X^2 + \sigma_Y^2 + 2\sigma_X\sigma_Y \leq 2(\sigma_X^2 + \sigma_Y^2)$ is true if and only if $\sigma_X^2 + \sigma_Y^2 \geq 2\sigma_X\sigma_Y$. I think there might be something in my answer that can be used to show that the latter assertion does always hold, and I will leave it to you to figure out what it might be. –  Dilip Sarwate Mar 9 '14 at 20:48