Let $A,B$ be two $n\times n$ matrices. It is known (see Simple Matrix inequality) that for $\beta\geq0$
$$ \frac{\beta}{1+\beta}AA^\ast-\beta BB^\ast\leq (A-B)(A-B)^\ast\leq 2(AA^*+BB^*). $$
What would be the lower bound of the $p$-th Shatten norm $$ \|(A-B)(A-B)^\ast\|_{S_{p/2}}^{1/2} $$ in terms of Shatten norm of $\|(AA^*+BB^*)\|_{S_q}$, for any possibility of $q$?
So far, I have obtained the following: For any $\beta\geq 0$, we have $$ (A-B)(A-B)^\ast\geq \frac{\beta}{1+\beta}AA^*-\beta BB^*. $$
We can write that
$$ \|(A-B)(A-B)^\ast\|_{S_{p/2}}^{1/2}\geq \|\frac{\beta}{1+\beta}AA^*-\beta BB^*\|_{S_{p/2}}^{1/2}\\ =\|\frac{\beta}{1+\beta}(AA^*+BB^*)-\frac{2\beta+\beta^2}{1+\beta} BB^*\|_{S_{p/2}}^{1/2} $$ Since for $x\geq y\geq0$, we have $\sqrt{x-y}\geq \sqrt x-\sqrt y$, we get that above line is $$ \geq \|\frac{\beta}{1+\beta}(AA^*+BB^*)\|_{S_{p/2}}^{1/2}- \|\frac{2\beta+\beta^2}{1+\beta} BB^*\|_{S_{p/2}}^{1/2}\\ \geq \left(\sqrt{\frac{\beta}{1+\beta}}-\sqrt{\frac{2\beta+\beta^2}{1+\beta}}\right)\|(AA^*+BB^*)^{1/2}\|_{S_{p}} $$ However, the coefficient in front of the norm is negative. So, one would just use a simple bound that $\|(AA^*+BB^*)\|_{S_q}\geq 0$ in this case.