Let $X$ and $Y$ be random variables in $L^2(\Omega,\mathcal{F},\mathbb{P})$ such that $$\mathbb{E}[X|Y] = f(Y)$$ almost-surely and $f:\mathbb{R}\to\mathbb{R}$ is non-increasing. Show that $Cov(X,Y)\leq 0$.
Tip: Show that $Cov(X,Y) = \mathbb{E}[(f(Y)-f(\mathbb{E}[Y]))(Y-\mathbb{E}[Y])]$.
We know that $$ \mathbb{E}[X] = \mathbb{E}[\mathbb{E}[X|Y]] = \mathbb{E}[f(Y)] \quad a.s. $$ thus \begin{align} Cov(X,Y) &= \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y]) \\ &= \mathbb{E}[\mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y]) \;| \; Y]] \\ &=\mathbb{E}[\mathbb{E}[(X-\mathbb{E}[X]) \;| \; Y](Y-\mathbb{E}[Y])] \end{align} which follows from the fact that $(Y-\mathbb{E}[Y])$ is $\sigma\langle Y\rangle$-measurable, \begin{align} &= \mathbb{E}[(f(Y)-\mathbb{E}[X])(Y-\mathbb{E}[Y])] \\ &= \mathbb{E}[(f(Y)-\mathbb{E}[f(Y)])(Y-\mathbb{E}[Y])] \\ \end{align}
however this is where I get stuck. I would appreciate some help.