Matrix $R$ is positive definite. Matrix $P$ is positive semidefinite. Matrix $D$ is some matrix with suitable size (need not to be square). All matrices are real and $D'$ denotes the transpose of $D$. $\geq$ denotes positive semidefinite.
How to prove the following?
$$ P-PD \left( R + D' P D \right)^{-1}D'P \geq 0$$
Edited
According to the idea given by @Karagounis Z
For any matrix $K$ we have
$$\left(\begin{matrix}R+K'K&K'\\K&I\end{matrix}\right) = \left(\begin{matrix}K'&R^{\frac{1}{2}}\\I&0\end{matrix}\right)\left(\begin{matrix}K&I\\R^{\frac{1}{2}}&0\end{matrix}\right)\geq0$$
Thus by the Schur complement (noting $R+K'K> 0$) we have,
$$I – K(R+K'K)^{-1}K'\geq 0$$
Thus by $P\geq 0$, there exists a unique positive semidefinite matrix $P^{\frac{1}{2}}\geq 0$ such that $P^{\frac{1}{2}}P^{\frac{1}{2}} = P$ and by letting $K = P^{\frac{1}{2}}D$
$$I-P^{\frac{1}{2}}D(R+D^TP^{\frac{1}{2}}P^{\frac{1}{2}}D)^{-1}D^TP^{\frac{1}{2}} \geq 0$$
Thus
$$I-P^{\frac{1}{2}}D(R+D^TPD)^{-1}D^TP^{\frac{1}{2}} \geq 0$$
Thus
$$P^{\frac{1}{2}}\left(I-P^{\frac{1}{2}}D(R+D^TPD)^{-1}D^TP^{\frac{1}{2}}\right) P^{\frac{1}{2}}\geq 0$$
Thus
$$ P-PD(R+D^TPD)^{-1}D^TP\geq 0$$
Or more directly, we consider
$$\left(\begin{matrix}R+D'PD & D'P\\PD&P\end{matrix}\right) = \left(\begin{matrix}D'P^{\frac{1}{2}} & R^{\frac{1}{2}}\\P^{\frac{1}{2}} & 0 \end{matrix}\right) \left(\begin{matrix}P^{\frac{1}{2}}D & P^{\frac{1}{2}}\\R^{\frac{1}{2}} & 0 \end{matrix}\right)\geq 0$$
Thus
$$ P-PD(R+D^TPD)^{-1}D^TP\geq 0$$
Best Answer
Note that $R + D'PD$ is positive definite, so by the Schur Complement Lemma, we have that the matrix $P - PD(R + D'PD)^{-1}D'P$ is positive semidefinite if and only if the block matrix
$$X=\left[\begin{array}{cc} R+D'PD & D'P \\ PD & P \end{array}\right]$$
is positive semidefinite. I think that $X$ can be written as $A'A$ for some matrix $A$ (try block matrices), which would show that it's positive semidefinite. Hope this helps.