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 $Tr[B^s (B R^2 B)^t] \ge Tr[B^s (R B^2 R)^t]$ ?
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Sign up to join this communityLet $B, R\in M_{n}(\mathbb{C})$ hermitian and $B$ positive semidefinite. Let $s,t \in \mathbb{R}$ and $s,t \ge 0$ .
Does then hold $Tr[B^s (B R^2 B)^t] \ge Tr[B^s (R B^2 R)^t]$ ?
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.