A related question is here.
Let $\Pi$ be a symmetric positive semidefinite projection i.e. $\Pi^2 = \Pi$ and $0\leq \Pi\leq I$ where the inequalities are meant in a positive semidefinite sense. Let $X$ and $A$ be arbitrary symmetric positive definite matrices. Is it true that
$$\text{tr}(X\Pi A\Pi)\leq \text{tr}(XA)?$$
I think it's false but I'm unable to construct a counterexample.
(working over $\mathbb R$)
For reasons of linearity, it suffices to check when $X$ is rank one.
Suppose $A$ is PSD with no zeros on its diagonal, $A\mathbf x =0$ for some $\mathbf x \neq \mathbf 0$ and $X := \mathbf x\mathbf x^T$. Finally, consider $\Pi$ as a diagonal matrix that is all zeros with a single one on its kth diagonal component. Then
$\text{trace}\big(X\Pi A\Pi\big) $
$= \text{trace}\big(\Pi\mathbf x \mathbf x^T\Pi A\big)$
$= x_k^2\text{trace}\big(\mathbf e_k\mathbf e_k^T A \big)$
$= x_k^2\cdot a_{k,k}$
$\gt 0 $
$= \text{trace}\big(A\mathbf x\mathbf x^T\big)$
$ = \text{trace}\big(XA\big)$