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.

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