I want to show that $\sqrt{A}B\sqrt{A}$ is operator concave with respect to $A$, when both $A$ and $B$ are Hermitian positive definite matrices.
I tested it numerically with different positive definite matrices, using the following condition: $$\sqrt{\frac{A+C}{2}}B\sqrt{\frac{A+C}{2}}>\frac{\sqrt{C}B\sqrt{C}}{2}+\frac{\sqrt{A}B\sqrt{A}}{2}$$
and it seems that it is true, but I don't know how to prove it.
Update: from an answer below this post, it turns out that $\sqrt{A}B\sqrt{A}$ is NOT operator concave.
The inequality does not always hold. If it holds in general, by passing $A$ and $C$ to the limit, it should hold (with $\ge$ in place of $>$) for every pair of positive semidefinite matrices $A,C$ and every positive definite $B$. Let $A=\operatorname{diag}(2,0),C=\operatorname{diag}(0,2)$ and $B=\pmatrix{2&-1\\ -1&2}$. The inequality then becomes $$ \pmatrix{2&-1\\ -1&2}\ge\pmatrix{0&0\\ 0&2}+\pmatrix{2&0\\ 0&0}=2I, $$ which is not true.