An operator inequality implying the usual and chaotic orders

Abstract

We prove that if positive invertible operators $A$ and $B$ satisfy an operator inequality $(B^{s/2}A^{(s-t)/2}B^tA^{(s-t)/2}B^{s/2})^{1\over{2s}}\geq B$ for some $t > s > 0$, then

(1) If $t \ge 3s-2 \ge 0$, then $\log B \geq \log A$, and if $t \ge s+2$ is additionally assumed, then $B \ge A$.

(2) If $s \in (0, 1/2)$, then $\log B \geq \log A$, and if $t \ge s+2$ is additionally assumed, then $B \ge A$.

It is an interesting application of the Furuta inequality. Furthermore we consider some related results.