TITLE: A conjectured trace inequality for some products of powers of matrices
QUESTION [7 upvotes]: Let $B, R\in M_{n}(\mathbb{C})$ hermitian and $B$ positive semidefinite.
Let $s,t \in \mathbb{R}$ and $s,t \ge 0$ . 
Does then hold $Tr[B^s (B R^2 B)^t] \ge Tr[B^s (R B^2 R)^t]$ ?

REPLY [5 votes]: I think the inequality is false. Consider for instance the choices
\begin{equation*}
B = \begin{bmatrix} 5&     6&    -2\\
     6  &  13  &   2\\
    -2  &   2  &   5\end{bmatrix},\quad
R = \begin{bmatrix}
 -8  &   4  &   4\\
     4 &   -2 &   -1\\
     4 &   -1 &    0   
    \end{bmatrix},\quad s=5,\ t=3.
\end{equation*}
Then, we have (computed using Mathematica)

lhs = Tr[MatrixPower[b, s].MatrixPower[b.r.r.b, t]]

which yields 415274500333934, whereas

rhs = Tr[MatrixPower[b, s].MatrixPower[r.b.b.r, t]]

yields 450223588494254, so that lhs-rhs = -34949088160320.