Simple Matrix inequality

Question

Let $A, B$ be two $n \times n$ matrices.

It is known that $(A-B)(A-B)^*\leq 2(AA^*+BB^*)$

How about the lower bound in terms of $AA^*$ and $BB^*$?

Answer 1

It is easy to obtain a lower bound by completing square. Pick any $\beta\ge0$ to obtain $$ \begin{aligned} &\quad\frac{1}{1+\beta}AA^\ast-AB^\ast-BA^\ast+(1+\beta)BB^\ast\\ &=\left(\frac{1}{\sqrt{1+\beta}}A-\sqrt{1+\beta}B\right)\left(\frac{1}{\sqrt{1+\beta}}A-\sqrt{1+\beta}B\right)^\ast\\ &\ge0 \end{aligned} $$ and hence $$ (A-B)(A-B)^\ast\ge\frac{\beta}{1+\beta}AA^\ast-\beta BB^\ast. $$