The following question kept me wondering for some time:


Given the symmetric matrices $A,B,C\in\mathbb{R}^{n×n}$ where $A$ and $C$ are positive definite (hence invertible), and $B$ is positive semidefinite (hence not necessarily invertible) with $\text{trace}(B)\neq0$, prove that


$\text{trace}\big\{C^{-1/2}BC^{−1/2}(A^{−1}+C^{−1/2}BC^{−1/2})^{−1}\big(A+\frac{n}{\text{trace}(B)}C\big)^{-1}\big\}\geq \text{trace}\big\{\big(A+\frac{n}{\text{trace}(B)}C\big)^{−2}A\big\}$


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


$\text{trace}\big\{B(A^{−1}+B)^{−1}\big(A+\frac{n}{\text{trace}(B)}I\big)^{-1}\big\}\geq \text{trace}\big\{\big(A+\frac{n}{\text{trace}(B)}I\big)^{−2}A\big\}$.


Please note that the matrix inversion lemma is not applicable at first to
$C^{-1/2}BC^{−1/2}(A^{−1}+C^{−1/2}BC^{−1/2})^{−1}$ since $B$ is positive semidefinite. Although I'm not sure, it seems like $\text{trace}(B)=\text{trace}{(\text{trace}(B)/n)I}$ should be utilized first in some way to arrive at some inequality to which the matrix inversion lemma can later be applied. I would highly appreciate if anyone can provide some help or suggestions on this.


Based on the suggestion of user2097, an invertible (positive definite) $B$ modifies the inequality to


$\text{trace}\big\{(A+C^{1/2}B^{-1}C^{1/2})^{−1}\big(A+\frac{n}{\text{trace}(B)}C\big)^{-1}A\big\}\geq \text{trace}\big\{\big(A+\frac{n}{\text{trace}(B)}C\big)^{−2}A\big\}$


which looks easier, but I was still unable to prove it. 



An inequality involving traces and matrix inversions
The last inequality is not true. For example, take $n=2$, $A, C$ to be identity matrix. It suffices to compare $\text{trace}\big\{(I+B^{-1})^{−1}\big\}$ and $\text{trace}\big\{\big(I+\frac{2}{\text{trace}(B)}I\big)^{−1}\big\}$. Now take $B=diag(1,2)$. You find that $\text{trace}\big\{(I+B^{-1})^{−1}\big\}=7/6<6/5=\text{trace}\big\{\big(I+\frac{2}{\text{trace}(B)}I\big)^{−1}\big\}$.


