Squares under Lowner order

Question

Suppose that $A, B$ are symmetric, positive definite, real matrices with the property $A \geq B$. Let $M$ be another symmetric, positive definite, real matrix (assumed to be of compatible dimensions). Is the following inequality true: $$ AMA \geq BMB $$ Of course, it is true for scalars, but I don't see an easy proof for matrices.

Answer 1

The answer is no, as seen from Example V.1.2, in Matrix Analysis, by Bhatia.

Reproduced below:

Suppose $$ A = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} \quad \mbox{and} \quad B = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} $$ Then note: $A - B = \left(\begin{smallmatrix} 1 & 0 \\ 0 & 0\end{smallmatrix}\right) \geq 0$. Additionally, it is easy to see that $B \geq 0$, and so $A \geq B \geq 0$. On the other hand, $$ A^2 = \begin{pmatrix} 5 & 3 \\ 3 & 2\end{pmatrix} \quad \mbox{and} \quad B^2 = \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix}. $$ Therefore, with $M = I$, $$ AMA - BMB = A^2 - B^2 = \begin{pmatrix} 3 & 1 \\ 1 & 0 \end{pmatrix} \not \geq 0, $$ since it has determinant $-1$.