Prove $\operatorname{Var}(X+Y) \le 2\operatorname{Var}(X) + 2\operatorname{Var}(Y)$ where $X$ and $Y$ are not necessarily independent I know how to show that $\operatorname{Var}(X+Y) = \operatorname{Var}(X) + \operatorname{Var}(Y) + 2\operatorname{Cov}(X,Y)$
I was also given the hint that I should use the triangle inequality to get
$|\operatorname{Var}(X) + \operatorname{Var}(Y) + 2\operatorname{Cov}(X,Y)| \le |\operatorname{Var}(x)| + |\operatorname{Var}(Y)| + |2\operatorname{Cov}(X,Y)|$
Honestly, I have no idea where to go from here. 

If you know how to show
$$\operatorname{Var}(X+Y) = \operatorname{Var}(X) + \operatorname{Var}(Y) + 2\operatorname{Cov}(X,Y)\tag1$$
then you can show
$$
\operatorname{Var}(X-Y) = \operatorname{Var}(X) + \operatorname{Var}(Y) - 2\operatorname{Cov}(X,Y).\tag2
$$
Now add equations (1) and (2) to get
$$
\operatorname{Var}(X+Y)+\operatorname{Var}(X-Y) = 2\operatorname{Var}(X) + 2\operatorname{Var}(Y).
$$
Finally the fact that variance of anything is non-negative implies that
$$
\operatorname{Var}(X+Y)\le \operatorname{Var}(X+Y)+\operatorname{Var}(X-Y).
$$
