Prove $M(M+I)^{-1}M \succeq \frac{1}{2}I$ for any PSD matrix $M \succeq I$

Question

For any $n \times n$ symmetrix matrices A, B, we define $A \succeq B$ if and only if $v^TAv \succeq v^TBv$ for all $v \in \mathbb{R}^n$

Answer 1

Since $M \succeq I$, $M$ is PD so it is invertible. We have $M^2 \succeq I$ and then $M \succeq M^{-1}$. From this and the fact that $M \succeq I$ we have $$2M \succeq M+I \succeq M^{-1}+I$$ The left and right sides of this inequality yield $$M \succeq \frac{1}{2}I+\frac{1}{2}M^{-1}$$ $$M \succeq \frac{1}{2}M^{-1}(M+I)$$ $$M(M+I)^{-1}M \succeq \frac{1}{2}I$$