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.

We have $$\operatorname{Var}[XY] = \operatorname{E}[X^2 Y^2] - \operatorname{E}[XY]^2.$$
Try to tackle each of these two terms by first conditioning on $X$ and then taking expectation.  For a full proof, hover below.

 Write $\operatorname{E}[XY] = \operatorname{E}[\operatorname{E}[XY \mid X]] = \operatorname{E}[X \operatorname{E}[Y \mid X] ] = \operatorname{E}[X]$ since $\operatorname{E}[Y \mid X] = 1$.  Lastly, we have $$\operatorname{E}[X^2Y^2] = \operatorname{E}[ X^2 \operatorname{E}[Y^2 \mid X]] \geq \operatorname{E}[ X^2 \operatorname{E}[Y\mid X]^2 ] = \operatorname{E}[X^2]$$ where we use Jensen's inequality for conditional expectation.  Thus $$\operatorname{Var}[XY] \geq \operatorname{E}[X^2] - \operatorname{E}[X]^2.$$ 

