Showing a matrix inequality

Question

Can someone help me with showing the following

Let $X$, $B$, and $C$ all be positive definite matrices. Show that the following inequality is true:

$$ (X + C)^{-1} (X B X' + C ) ( X + C)^{-1} \ge (B^{-1} + C)^{-1}$$

for all $X$, where $A \ge B$ means $x' A x \ge x' B x$ for all column vector $x$.

I can show it for the 1-dimensional analogue, but I'm lost on what to do in the matrix space.

Thanks

Answer 1

Here is a simpler proof. Consider the matrix \begin{align} A &=\pmatrix{XBX^T+C&X+C\\ X^T+C&B^{-1}+C}\\ &=\pmatrix{XB^{1/2}\\ B^{-1/2}}\pmatrix{B^{1/2}X^T&B^{-1/2}}+\pmatrix{C^{1/2}\\ C^{1/2}}\pmatrix{C^{1/2}&C^{1/2}}. \end{align} $A$ is the sum of two Gram matrices. Hence it is positive semidefinite. Since $B^{-1}+C$ is positive definite, its Schur complement in $A$, i.e. the matrix $(XBX^T+C)-(X+C)(B^{-1}+C)^{-1}(X^T+C)$, must be positive semidefinite. Therefore $$ (X+C)^{-1}(XBX^T+C)(X^T+C)^{-1}\ge (B^{-1}+C)^{-1}. $$ as long as $X+C$ is invertible.

Answer 2

Put $D = B^{1/2}CB^{1/2}$ and $Y = B^{1/2}XB^{1/2}$, the inequality becomes \begin{align} (X+C)^{-1}(XBX+C)(X+C)^{-1} &\ge (B^{-1}+C)^{-1},\\ B^{1/2}(Y+D)^{-1}(Y^2+D)(Y+D)^{-1}B^{1/2} &\ge (B^{-1}+B^{-1/2}D^{-1}B^{-1/2})^{-1},\\ (Y+D)^{-1}(Y^2+D)(Y+D)^{-1} &\ge (I+D^{-1})^{-1},\\ (Y^2 + D) &\ge (Y+D)(I+D)^{-1}(Y+D). \end{align} Left and right multiply both sides by $D^{-1/2}$, the inequality is equivalent to \begin{align} D^{-1/2}(Y^2 + D)D^{-1/2} &\ge D^{-1/2}(Y+D)(I+D)^{-1}(Y+D)D^{-1/2}\\ &= D^{-1/2}(Y+D)D^{-1/2}(D^{-1}+I)^{-1}D^{-1/2}(Y+D)D^{-1/2}. \end{align} Put $Z = D^{1/2}Y^{-1}D^{1/2}$, we get $Y = D^{1/2}Z^{-1}D^{1/2}$ and \begin{align} Z^{-1}DZ^{-1}+I &\ge (Z^{-1}+I)(D^{-1}+I)^{-1}(Z^{-1}+I),\\ D+Z^2 &\ge (I+Z)(D^{-1}+I)^{-1}(I+Z),\\ &= (I+Z)^2 - (I+Z) (I+D)^{-1} (I+Z),\\ D + (I+Z) (I+D)^{-1} (I+Z) &\ge 2Z + I. \end{align} Put $E = I+D$ and $W=Z+I$, the inequality further reduces to \begin{align} E + WE^{-1}W &\ge 2W,\\ E^{1/2}(I-E^{-1/2}WE^{-1/2})^2E^{1/2} &\ge 0, \end{align} which is true because both $E^{1/2}$ and $I-E^{-1/2}WE^{-1/2}$ are Hermitian. Note that in the course of proof, we do not need the positive definiteness of $X$. The inequality still holds if $X$ is only Hermitian and $X+C$ is invertible.