$A\geq B\geq 0$ assures $(B^ rA^ pB^ r)^ {1/q}\geq B^ {(p+2r)/q}$ for $r\geq 0$, $p\geq 0$, $q\geq 1$ with $(1+2r)q\geq p+2r$

Abstract:

An operator means a bounded linear operator on a Hilbert space. This paper proves the assertion made in its title. Theorem 1 yields the famous result that $A \geq B \geq 0$ assures ${A^\alpha } \geq {B^\alpha }$ for each $\alpha \in [0,1]$ when we put $r = 0$ in Theorem 1. Also Corollary 1 implies that $A \geq B \geq 0$ assures ${(B{A^p}B)^{1/p}} \geq {B^{(p + 2)/p}}$ for each $p \geq 1$ and this inequality for $p = 2$ is just an affirmative answer to a conjecture posed by Chan and Kwong. We cite three counterexamples related to Theorem 1 and Corollary 1.