Trace inequality for square root of product of matrices All the matrices in this question are symmetric positive semidefinite. The notation $A\leq B$ is used to denote that $B-A$ is positive semidefinite.
Suppose $0\leq \Pi\leq I$. Let $X\geq 0$ be some positive semidefinite matrix. Is it true that
$$\text{Tr}\left(X^{1/2}\right) \geq \text{Tr}\left((\Pi X\Pi)^{1/2}\right)$$
Note that $\Pi X\Pi$ is symmetric and hence this product is also positive-semidefinite and the square root is thus well defined.
In the answer linked above, it is not clear how to "open up" the term $(\Pi X\Pi)^{1/2}$. I suspect that knowing this might help me prove the inequality (if it is true).
 A: Proof 1. In general, if $A$ and $B$ are two rectangular matrix such that $AB$ is a square matrix, then $\sigma_i(AB)\le\sigma_i(A)\sigma_1(B)$. Now put $A=X^{1/2}$ and $P=\Pi$, we obtain
\begin{aligned}
\operatorname{tr}\left((\Pi X\Pi)^{1/2}\right)
&=\operatorname{tr}\left(\left((X^{1/2}\Pi)^T (X^{1/2}\Pi)\right)^{1/2}\right)\\
&=\sum_i\sigma_i\left(X^{1/2}\Pi\right)\\
&\le\sum_i\sigma_i\left(X^{1/2}\right)\sigma_1(\Pi)\\
&\le\sum_i\sigma_i\left(X^{1/2}\right)
=\operatorname{tr}(X^{1/2}).
\end{aligned}
Proof 2. By a continuity argument and by a change of orthonormal basis, we may assume that $\Pi$ is a positive diagonal matrix. Let $Y=\Pi^{-1}(\Pi X\Pi)^{1/2}\Pi^{-1}$ and $A=Y\,\Pi$. Then
\begin{aligned}
X&=Y\,\Pi^2Y=AA^T,\\
(\Pi X\Pi)^{1/2}&=\Pi\,Y\,\Pi=\Pi\,A.
\end{aligned}
Note that $A=Y\,\Pi$ has a nonnegative diagonal because $Y$ is positive semidefinite and $\Pi$ is a positive diagonal matrix. Therefore $\operatorname{tr}(A)\ge\operatorname{tr}(\Pi\,A)$ and in turn,
$$
\operatorname{tr}(X^{1/2})
=\operatorname{tr}\left((AA^T)^{1/2}\right)
=\sum_i\sigma_i(A)
\ge\operatorname{tr}(A)
\ge\operatorname{tr}(\Pi\,A)
=\operatorname{tr}\left((\Pi X\Pi)^{1/2}\right).
$$
