Trace inequality on positive matrices

Question

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?

Answer 1

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 $B^\ast AB\geq 0$).

Answer 2

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.

Answer 3

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}