Trace inequality on positive matrices Let $A,B,C\geq 0$ be self-adjoint matrices.
Assume $A\leq B$. Is it true that 
$$\mathrm{tr}(ACAC) \leq \mathrm{tr}(BCBC)?$$
How to prove this?
 A: Just for other users with a similar question, here is a direct solution:
\begin{align}\mathrm{tr}(BCBC)-\mathrm{tr}(ACAC)&=\mathrm{tr}(BCBC-ACAC+BCAC-ACBC)\\
&=\mathrm{tr}((B-A)C(B+A)C)\\
&=\mathrm{tr}((B-A)^{\frac 12}C(B+A)C(B-A)^{\frac 1 2})\\
&\geq 0.
\end{align}
Observe that in the first line the arguments on both sides differ only by a commutator, so their trace is the same. In the third line I just rearranged the factors to get a positive argument (in general, $A\geq 0$ implies $B^\ast AB\geq 0$).
A: I think I solved it. It follows from the fact that if $A\geq B$, then $\mathrm{tr}AC \geq \mathrm{tr} BC$ for any $C\geq 0$. This, however, follows from this question.
A: From $B\ge A\ge0$ and $C\ge0$, we get $Y=C^{1/2}BC^{1/2}\ge X=C^{1/2}AC^{1/2}\ge0$. Let $Z=Y-X\ge 0$. Then
\begin{align}
\operatorname{tr}(BCBC)&=\operatorname{tr}\left((C^{1/2}BC^{1/2})^2\right)\\
&=\operatorname{tr}(Y^2)\\
&=\operatorname{tr}(X^2+XZ+ZX+Z^2)\\
&=\operatorname{tr}(X^2)+2\operatorname{tr}(Z^{1/2}XZ^{1/2})+\operatorname{tr}(Z^2)\\
&\ge\operatorname{tr}(X^2)\\
&=\operatorname{tr}\left((C^{1/2}AC^{1/2})^2\right)\\
&=\operatorname{tr}(ACAC).
\end{align}
