Does $A^2 \geq B^2 > 0$ imply $ACA \geq BCB$ for square positive definite matrices?
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 $ACA\ge BCB$.
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