Suppose that $A, B$ are Hermitian positive definite matrices of the same order and $0\le p\le 1$. Using a standard approach in matrix analysis, one can show that $\|A^{1-p}B^p\|\ge \|A\sharp_p B\|$, where $A\sharp_p B:=A^{1/2}(A^{-1/2}BA^{-1/2})^pB^{1/2}$ which is sometimes called the weighted geometric mean and has a geometric interpretation. The matrix norm here is spectral norm (i.e., largest singular value).

I tried to play with $\|A^{1-p}B^p\|\ge \|A\sharp_p B\|$ a little bit, a question suddenly occured to me: Is it true $$\|AB\|\ge \|(A\sharp_p B)(A\sharp_{1-p}B)\|?$$

I ran some simulations yet no counterexample showed up... the standard approach I know seems not work, so I am looking for some new ingredients.

Suppose that $A, B$ are Hermitian positive definite matrices of the same order and $0\le p\le 1$. Using a standard approach in matrix analysis, one can show that $\|A^{1-p}B^p\|\ge \|A\sharp_p B\|$, where $A\sharp_p B:=A^{1/2}(A^{-1/2}BA^{-1/2})^pA^{1/2}$ which is sometimes called the weighted geometric mean and has a geometric interpretation. The matrix norm here is spectral norm (i.e., largest singular value).

I tried to play with $\|A^{1-p}B^p\|\ge \|A\sharp_p B\|$ a little bit, a question suddenly occured to me: Is it true $$\|AB\|\ge \|(A\sharp_p B)(A\sharp_{1-p}B)\|?$$

I ran some simulations yet no counterexample showed up... the standard approach I know seems not work, so I am looking for some new ingredients.