Question: 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.

Answer: 
I'll write a proof for the (easy) case $p=1/2$. The general case seems more tricky.

For $p=1/2$, the task is to prove $\| (A\sharp B)^2\| \le \|AB\|$, or equivalently that $\|A\sharp B\|^2 \le \|AB\|$. But we know that $\|A\sharp B\| \le \|A^{1/2}B^{1/2}\|$, so if we show that $\|A^{1/2}B^{1/2}\|^2 \le \|AB\|$ we'll be done. 

The latter inequality follows from the $k=1,r=2$ case of the well-known *log-majorization*:
\begin{equation\*}
 \prod\_{i=1}^k \lambda^r(A^{1/2}BA^{1/2}) \le \prod\_{i=1}^k \lambda(A^{r/2}B^rA^{r/2}),\qquad k=1,\ldots,n, r \ge 1,
\end{equation\*}
for $n\times n$ positive definite matrices $A$ and $B$.
Her is an elementary proof of the case $p=\frac12$. As noted by Suvrit, it is enough to prove $\|A^{1/2}B^{1/2}\|^2 \le \|AB\|$, or equivalently $\|HK\|^2\le\|H^2K^2\|$. This is true because of
$$\|HK\|^2=\rho((HK)^\*HK)=\rho(KH^2K)=\rho(H^2K^2)\le\|H^2K^2\|.$$
Hereabove, $\rho$ is th spectral radius and the last inequality is true for every subordinated norm.