# Inequality involving traces and matrix inversions

The following question kept me wondering for some weeks:

Given the symmetric matrices $A,B,C\in\mathbb{R}^{n\times n}$ where $A$ and $C$ are positive definite (hence invertible), and $B$ is positive semidefinite (hence not necessarily invertible) with $\operatorname{trace}(B)\neq 0$, prove that

$$\operatorname{trace}\left\{C^{-1/2}BC^{-1/2}(A^{-1}+C^{-1/2}BC^{-1/2})^{-1}(A+\frac{n}{\operatorname{trace}(B)}C)^{-1}\right\}\geq \operatorname{trace}\left\{(A+\frac{n}{\operatorname{trace}(B)}C)^{-2}A \right\}.$$

If it would help, one can also consider the simpler version with $C=I$: prove that

$$\operatorname{trace}\left\{B(A^{-1}+B)^{-1}(A+\frac{n}{\operatorname{trace}(B)}I)^{-1}\right\}\geq \operatorname{trace}\left\{(A+\frac{n}{\operatorname{trace}(B)}I)^{-2}A \right\}.$$

Please note that the matrix inversion lemma is not applicable at first as $B$ is positive semidefinite. Although I'm not sure, it seems like $\operatorname{trace}(B)=\operatorname{trace}\{(\operatorname{trace}(B)/n)I\}$ should be utilized in some way, and it may also help to interpret the trace operator in terms of the Frobenius norm. I would highly appreciate if anyone can provide some help or suggestions on this.

• Crossposting: mathoverflow.net/questions/178089/…. Aug 8, 2014 at 11:55
• This inequality was shown to be incorrect at Mathoverflow. Therefore, people can disregard my question. Aug 16, 2014 at 22:28

M. Lin's answer at Math Overflow show that the inequality doesn't need to hold even in the case $C=I$.