Suppose $\mathbf{A}\in\mathbb{C}^{M \times M}$, and diagonal matrix $\mathbf{B}\in\mathbb{C}^{M \times M}$. $\mathbf{A}\succeq \mathbf{0}$, and $\mathbf{0}\preceq\mathbf{B}\preceq\mathbf{I}$.
How to prove $$ (\mathbf{I}+\mathbf{A})^{-1} \succeq \mathbf{B}(\mathbf{I}+\mathbf{A}\mathbf{B})^{-1} $$ When $\mathbf{0}\prec\mathbf{B}\preceq\mathbf{I}$, this equation is easy to prove, but how to prove when $\mathbf{0}\preceq\mathbf{B}\preceq\mathbf{I}$? Thanks.
$\mathbf{B}$ can be expressed as $\mathbf{B}=diag\{b_1,b_2,...,b_M\}$, where $0\leq b_m \leq 1$ is the diagonal elements of $\mathbf{B}$. If we can construct $\mathbf{B}_t =\mathbf{B}+t\mathbf{K}$ for $0 < t \leq 1$ and diagonal matrix $\mathbf{K} \succeq \mathbf{0}$ satisfying $\mathbf{0}\prec\mathbf{B}_t\preceq\mathbf{I}$, then we can proof $$ (\mathbf{I}+\mathbf{A})^{-1} \succeq \mathbf{B}_t(\mathbf{I}+\mathbf{A}\mathbf{B}_t)^{-1} $$ When $t\to0^{+}$, we can have $(\mathbf{I}+\mathbf{A})^{-1} \succeq \mathbf{B}(\mathbf{I}+\mathbf{A}\mathbf{B})^{-1}$. But how to design the diagonal matrix $\mathbf{K}$?
