Expectation of conditional variance (Exercise 4.6.7 from Grimmett and Stirzaker): Let $X$ and $Y$ be random variables with correlation $\rho$. Show that $E(\mathrm{var}(Y|X)) \leq (1 - \rho^2) \mathrm{var}(Y)$.
This is how far I got:
By definition,
$$ \mathrm{var}(Y|X) = E(Y^2|X) - E(Y|X)^2. $$
Taking expectations of both sides gives
\begin{align} E(\mathrm{var}(Y|X)) &= E(E(Y^2|X)) - E(E(Y|X)^2), \\ &= E(Y^2) - E(E(Y|X)^2). \end{align}
And from the Cauchy-Schwarz inequality we have \begin{align} E(XY)^2 &= E(X E(Y|X))^2 \\ &\leq E(X^2)E(E(Y|X)^2), \\ \implies E(E(Y|X)^2) &\geq \frac{E(XY)^2}{E(X^2)}. \end{align}
So therefore, $$ E(\mathrm{var}(Y|X)) \leq E(Y^2) - \frac{E(XY)^2}{E(X^2)}. $$
How can I proceed from here? I tried expanding $(1 - \rho^2) \mathrm{var}(Y)$ after substituting $$ \rho^2 = \frac{(E(XY) - E(X)E(Y))^2}{(E(Y^2) - E(Y)^2)(E(X^2) - E(X)^2)} $$ but couldn't get the right hand side to match the left hand side.