Trace of positive semidefinite matrix

Question

Let $A =X +B$ with $X= (X_1+X_2) $all three semidefinite positive and B definite positive. How can i proove that $$ trace A^{-1}X \leq trace( (X_1+ B)^{-1}X_1 +(X_2 +B)^{-1}X_2)$$

Answer 1

Since $A \ge X_i+B$, we have $A^{-1} \le (X_i+B)^{-1}$ and in turn $X_i^{1/2}A^{-1}X_i^{1/2} \le X_i^{1/2}(X_i+B)^{-1}X_i^{1/2}$. Thus $\operatorname{tr}(A^{-1}X_i) \le \operatorname{tr}\left((X_i+B)^{-1}X_i\right)$. Sum up each side for $i=1,2$, the result follows.

Answer 2

Let $M_1=A^{-1}X_1$, $N_1=(X_1+B)^{-1}X_1$, and $Q_1=(X_1+B)^{-1}-A^{-1}$. Because $A-(X_1+B)=X_2$ is positive semidefinite, so is $Q_1$. Then, \begin{align*} \operatorname{Tr}(N_1-M_1)=\operatorname{Tr}\Big[Q_1X_1\Big]=\operatorname{Tr}\Big[Q_1^{1/2}Q_1^{1/2}X_1\Big]=\operatorname{Tr}\Big[Q_1^{1/2}X_1Q_1^{1/2}\Big]\geq0. \end{align*} Then, you can do the same with $M_2=A^{-1}X_2$, $N_2=(X_2+B)^{-1}X_2$, and $Q_2=(X_2+B)^{-1}-A^{-1}$.