Does $A^2 \geq B^2 > 0$ imply $ACA \geq BCB$ for square positive definite matrices? 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!
 A: 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
A: I don't know if there are any nice (i.e. not-too-strong) conditions for the inequality to hold, but I'm sure that it doesn't always hold, even when $C$ is positive definite. Counterexample:
\begin{align}
A&=A^2=I,\\
B&=B^2=\frac12\pmatrix{1&1\\ 1&1},\\
C&=\operatorname{diag}(1,4).
\end{align}
In this case, we have $A^2\ge B^2\ge 0$ but $ACA-BCB=\frac14\pmatrix{-1&-5\\ -5&11}$ is indefinite. While $B$ is not positive definite here, by continuity, we can obtain a valid counterexample by adding a small positive multiple of $I$ to both $A$ and $B$.
Edit.


*

*Note that if $ACA\ge BCB$ for all real symmetric $C$, we must have $A=B$ because $AIA\ge BIB$ and $A(-I)A\ge B(-I)B$ imply that $A^2=B^2$.

*It isn't quite meaningful to consider $ACA\ge BCB$ for all $C\ge0$ either. In particular, if $A(vv^\ast)A\ge B(vv^\ast)B$ for every vector $v$, then $Bv$ must be equal to $\lambda_v Av$ for some $0\le\lambda_v\le1$. Therefore, by linearity, $B=\lambda A$ for some $0\le\lambda\le1$.

*It is interesting to ask, however, if $A\ge B>0$ and $A^2\ge B^2$, what class of $C$ (under perhaps some additional conditions on $A$ and $B$) would satisfy the inequality $ACA\ge BCB$.

A: Here I posted a related question. My (preliminary) conclusion to this question might be a valuable idea/suggestion to your answer: 
I strongly assume (!) when your matrices commute (i.e. $(AB)^T AB = AB (AB)^T$) then $ACA \geq BCB$. This could be the case when $A$, $B$, and $C$ are diagonal matrices (and of course in the trivial cases of $C$ or $A$ and $B$ being multiples of $I$).
