Given that $A=\text{diag}(a_1,\dots,a_n)$ is a diagonal matrix (this we fix) with positive values and $B=\text{diag}(b_1,\dots,b_n)$ is a another diagonal matrix (this we get to choose) with positive values, I know that
\begin{align*}
\text{Tr} \left( A B \left( B + A^{-1} \right)^{-1} \right)
&= \sum_{i=1}^n \frac{a_i^2}{a_i+b_i^{-1}}
\end{align*}
is maximised when $b_i$'s (these are also the eigenvalues) are maximized, which further imply that $\text{Tr}(B)$ (these are also the sum of eigenvales) is maximized. Is this still true if $B$ is just a general positive definite matrix (i.e., Is $\text{Tr}(AB(B+A^{-1})^{-1})$ maximized when $\text{Tr}(B)$ is maximized — note that $A$ remains fixed)? (Note that $A$ is still a diagonal matrix with positive values.)
Best Answer
The observation that I think you have in mind is as follows. For diagonal matrices $B = \operatorname{diag}(b_1,\dots,b_n)$ and $C = \operatorname{diag}(c_1,\dots,c_n)$ and the function $$ f(B) = \operatorname{tr}(AB(B + A)^{-1}), $$ it holds that if $b_i \geq c_i > 0$ for all $i$, then $f(B) \geq f(C)$. One natural extension of this property would be that for positive definite matrices $B\succeq C$ (i.e. $B - C$ is positive semidefinite), it holds that $f(B) \geq f(C)$. As it turns out, this result does hold.
First, note that we can rewrite $$ \begin{align} AB(B + A^{-1})^{-1} &= AB(A^{-1}[A + B^{-1}]B)^{-1} \\ & = ABB^{-1}(A + B^{-1})^{-1}A \\ &= A(A + B^{-1})^{-1}A. \end{align} $$ Now, it's easier to see that if $B$ is positive definite, then this expression (namely $A(A + B^{-1})^{-1}A$) is positive definite. Note that $$ B \succeq C \implies\\ B^{-1} \preceq C^{-1} \implies\\ A + B^{-1} \preceq A + C^{-1} \implies\\ (A + B^{-1})^{-1} \succeq (A + C^{-1})^{-1} \implies\\ A(A + B^{-1})^{-1}A \succeq A(A + C^{-1})^{-1}A. $$ It follows that $$ f(B) = \operatorname{tr}[A(A + B^{-1})^{-1}A] \geq \operatorname{tr}[A(A + C^{-1})^{-1}A] = f(C), $$ which is what we wanted to show.