TITLE: Is there a trace inequality for the product of a sequence of hermitian postive definite matrices?
QUESTION [6 upvotes]: Let $A$ and $B$ be two Hermitian matrices with positive eigenvalues. 
Let $k>0$ be a integer.
Let $P=(P_1,P_2,\dots,P_{2k})$ be a sequence of $k$ $A$s and $k$ $B$s in any given order.
Do we have
${\rm tr}\,\prod^{2k}_{i=1} P_i \leq {\rm tr}\,A^k B^k$ ?

REPLY [5 votes]: A counterexample was apparently first found by Plevnik (2016) (pdf):
$$
A = \begin{bmatrix}76&0&0\\0&0&0\\0&0&1\end{bmatrix}
\qquad\text{and}\qquad
B = \begin{bmatrix}20&-14&13\\-14&2880&3100\\13&3100&3380\end{bmatrix}
$$
give
$$
\operatorname{tr} A^4BAB^4 = 7608677695167720100 > 7566365725138281700 = \operatorname{tr} A^5B^5.
$$
To get one with $A$ positive definite as requested, replace entry $A_{22}$ by $0.01$: the two traces become
$$
7584680876077508226.18611992\ \ > \ \ 7566365725573314229.03610008.
$$