Show that $2I - (MX-S)^t(S^2 + MX^2M^t)^{-1}(MX - S)$ is positive semidefinite

Question

Show that $$2I - (MX-S)^t(S^2 + MX^2M^t)^{-1}(MX - S)$$ where $X$ and $S$ are positive diagonal matrices, is a positive semidefinite (PSD) matrix.

I can say $(S^2 + MX^2M^t)^{-1}$ is a PD (positive definite) matrix and it is symmetric also. But after that I can not say anything. If it is $2I +$(expression) then it is clear.

Answer 1

According to the Schur complement lemma, your inequality is equivalent to proving that $$\left( \begin{matrix} 2I & S-MX \\ S-XM^T & S^2+MX^2M^T \\ \end{matrix}\right)\ge 0 $$ which is easy.