Given that $\operatorname{E}[Y\mid X] = 1$, show that $\operatorname{Var}(XY)\geqslant\operatorname{Var}(X)$ 
Given that $E[Y\mid X] = 1$, show that $\operatorname{Var}(XY)\geqslant\operatorname{Var}(X)$.

So I tried to expand $\operatorname{Var}(XY) = \operatorname{E}(X^2 Y^2) - 1$ and was stuck here.

$\newcommand{\v}{\operatorname{var}}\newcommand{\e}{\operatorname{E}}$By the law of total variance, you have
\begin{align}
\v(XY) & = \v(\e(XY\mid X)) + \e(\v(XY\mid X)) \\[10pt]
& = \v(X\e(Y\mid X)) + \e(X^2\v(Y\mid X)) \\[10pt]
& = \v(X) + \e(X^2\v(Y\mid X)) \\[10pt]
& \ge \v(X).
\end{align}
