Loewner ordering of symmetric positive definite matrices after transformation

Question

If we consider two symmetric positive definite matrices $A$ and $B$ with the ordering $A>B$, i.e., $A-B>0$ is positive (semi-)definite, does then

$B^{-1} A B^{-1} > A$

hold? It can be shown that

$B^{-1}AB^{-1}>B^{-1}>A^{-1}$ holds because $A>B$ is equivalent to $A^{-1}

Does anyone know whether also $B^{-1} A B^{-1} > A$ holds?

Answer 1

No. Try $B=2I$ and $A=5I$. Then $A \succeq B$, but $B^{-1}AB^{-1}=(1/4)A \not{\succeq} A$.