TITLE: Bound the variance of the product of two random varables.
QUESTION [7 upvotes]: For two random variables $X$ and $Y$ show that the following inequality holds
$$\mathrm{Var}(XY)\leq 2\|Y\|_{\infty}^{2}\mathrm{Var}(X)+2\|X\|_{\infty}^{2}\mathrm{Var}(Y).$$
Well first I tried to show it for just indicators functions by I couldn't even show that. 
Any tips?

REPLY [8 votes]: If $U$ and $V$ are two random variables, then the inequality 
$$\mathrm{Var}(U+V)\leqslant 2\mathrm{Var}(U)+2\mathrm{Var}(V)$$
takes place. Using this inequality with $U:=(X-\mathbb E[X])Y$ and $V:=\mathbb E[X]Y$, we obtain that 
$$\mathrm{Var}(XY)\leqslant 2\mathrm{Var}((X-\mathbb E[X])Y)+2\mathrm{Var}(\mathbb E[X]Y).$$
Since 
\begin{align}\mathrm{Var}((X-\mathbb E[X])Y)&=\mathbb E\left[((X-\mathbb E[X])Y)^2\right]-
\left(\mathbb E\left[(X-\mathbb E[X])Y\right]\right)^2\\
&\leqslant \mathbb E\left[((X-\mathbb E[X])Y)^2\right]\\
&\leqslant \mathbb E\left[(X-\mathbb E[X])^2\right]\lVert Y\rVert_\infty^2\\
&=\mathrm{Var}(X)\lVert Y\rVert_\infty^2,
\end{align}
we have showed the inequality 
$$\mathrm{Var}(XY)\leqslant 2\mathrm{Var}(X)\lVert Y\rVert_\infty^2+2(\mathbb E[X])^2\mathrm{Var}(Y).$$