TITLE: $A,B$ be Hermitian.Is this true that $tr[(AB)^2]\le tr(A^2B^2)$?
QUESTION [7 upvotes]: Suppose $A,B \in {M_n}$ be Hermitian.Is this true that $tr[(AB)^2]\le tr(A^2B^2)$?

REPLY [15 votes]: Yes.  Equality holds iff $AB = BA$.
Hint: Note that $AB - BA$ is skew-Hermitian, and that
$$
2\operatorname{trace}[(AB)^2] - 2\operatorname{trace}(A^2B^2) =\\
\operatorname{trace}(ABAB + BABA -ABBA - BAAB)=\\
\operatorname{trace}[(AB - BA)^2]
$$
Note: The inequality assumes that both $\operatorname{trace}[(AB)^2]$ and $\operatorname{trace}(A^2B^2)$ are real.  This is fine because the product of Hermitian matrices necessarily has a real trace.  Note that
$$
\operatorname{trace}[(A+aI)(B+bI)] =\\
\operatorname{trace}(AB) + 
b\operatorname{trace}(A) + 
a \operatorname{trace}(B) +
abn
$$
and that $a,b>0$ may be chosen so that $A +aI$ and $B+bI$ are positive definite.  The product of (Hermitian) positive definite matrices always has positive eigenvalues.