Let $A$ be matrix in $M\_{n}$ (i.e., $n\times n$ complex matrices), and $\|A\|\le 1$, we call it a contraction.


Assume that $A$ and $B$ are contractions such that
$I-AA^\*$ and $I-BB^\*$ are positive-definite. 


How to show that
$$\text{Tr}\left(1-AA^\*\right)^{-1}\cdot \text{Tr}\left(1-BB^\*\right)^{-1} \ge \left(\text{Tr}(1-AB^\*)^{-1}\right)^2.$$




---


This problem is similar to the Cauchy-Schwarz inequality. (if you don't know the contraction matrix) and I found this paper give the contraction. <http://ac.els-cdn.com/S0024379507003710/1-s2.0-S0024379507003710-main.pdf?_tid=b1696360-f73a-11e3-9613-00000aab0f01&acdnat=1403131900_0ced864818647681ff916fc376a0f461>



A similar Cauchy-Schwarz inequality with linear-algebra
Here's one way to prove this.


Assuming $A$ and $B$ are contractions, first recall *Hua's Matrix Identity*
\begin{equation\*}
 (I-B^\*B) + (A-B)^\*(I-AA^\*)^{-1}(A-B) = (I-B^\*A)(I-A^\*A)^{-1}(I-A^\*B).
\end{equation\*}
From this it identity immediately follows that the following matrix is (Hermitian) positive definite:
\begin{equation\*}
 \begin{pmatrix}
 (I-A^\*A)^{-1} & (I-B^\*A)^{-1}\\
 (I-A^\*B)^{-1} & (I-B^\*B)^{-1}
 \end{pmatrix} \succeq 0.
\end{equation\*}
From the operator inequality, taking trace (which preserves the positive semidefiniteness), we obtain the inequality in the OP.


