TITLE: Does $A^2 \geq B^2 > 0$ imply $ACA \geq BCB$ for square positive definite matrices?
QUESTION [8 upvotes]: Assume we have two $n \times n$ real nondegenerate matrices $ A^2 $ and $B^2$, such that
$$
A^2 \geq B^2 > 0,
$$
where "$\geq$" means positive semidefinite (Loewner) ordering. Does the following inequality holds for any real matrix $C$ 
$$
ACA \geq BCB \ ?
$$
If not, under which conditions on $C$ (or additional conditions on $A$ and $B$) does it holds? 
I would appreciate any ideas, suggestions, counterexamples.
Thanks!

REPLY [4 votes]: In [Theorem 2.6a, 1] it is proven using Fuglede-Putnam that

Proposition. For $I \geq A, B \geq 0$ we have $\sqrt{A} B \sqrt{A} \leq B$ if and only if $AB = BA$.

(The matrix $\sqrt{A} B \sqrt{A}$ models the sequential measurement of first $A$ andthen $B$ in quantum mechanics and its properties are studied by various authors.  By them it is called the "sequential product").
As a corollary we can partially answer your question:

Corollary.  Suppose $I \geq A,B \geq 0$ with $A^2 \geq B^2$.  If for all real $C$, we have $ACA \geq BCB$, then $AB=BA$.

Proof. Consider $C:=A$. By assumption $A \geq A^2 \geq BAB=\sqrt{B^2}A\sqrt{B^2}$ and so $B^2$ commutes with $A$.  But then also~$B$ commutes with $A$. QED
[1] Sequential quantum measurements by Gudder and Nagy