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 $Tr[B^s (B R^2 B)^t] \ge Tr[B^s (R B^2 R)^t]$ ?



A conjectured trace inequality for some products of powers of matrices
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**.


