Given three positive operators $R,S,T\in B(H)$, with $\left|\left|T\right|\right|\leq 1$ and $\left|\left|S\right|\right|\leq 1$, I would like to know whether the inequality
$\left|\left|SRS\right|\right|\leq \left|\left|(S+T)R(S+T)\right|\right|$
holds. I would be equally happy if the right hand side is replaced by a polynomial of the original right hand side.
I could achieve
$\left|\left|SRS\right|\right|\leq \left|\left|(S+T)^{1/2}R(S+T)^{1/2}\right|\right|$,
mainly by using the identity $||x^*x||=||x||^2$ for bounded operators on a Hilbert space, but it's not good enough for me.
Thanks for any help!
The polynomial version of the question, I wouldn't know how to tackle. But the inequality $$ \|SRS\|\leq\|(S+T)R(S+T)\| $$ can fail, even when $0\leq R\leq 1$.
Let $$ S=\begin{bmatrix} 1&0\\0&0\end{bmatrix},\qquad T=\begin{bmatrix} 1&1\\1&1\end{bmatrix},\qquad R=vv^T, $$ where $$ v=\begin{bmatrix} -\sqrt{\frac{5-\sqrt5}{10}}\\ \sqrt{\frac{5+\sqrt5}{10}}\end{bmatrix}. $$ As $v$ is a unit vector, $R$ is a projection. We have $$ SRS=R_{11}=(v_1)^2=\frac{5-\sqrt5}{10}\,S,\qquad \text{ so }\qquad \|SRS\|=\frac{5-\sqrt5}{10}\simeq 0.276. $$ Meanwhile, $$ \begin{bmatrix} 2&1\\1&1\end{bmatrix}v=\frac{3-\sqrt5}2\,v $$ so $$ (S+T)R(S+T)=\begin{bmatrix} 2&1\\1&1\end{bmatrix}v\Bigg(\begin{bmatrix} 2&1\\1&1\end{bmatrix}v\Bigg)^T =\Big(\frac{3-\sqrt5}2\Big)^2\,R. $$ As $R$ is a projection, $$ \|(S+T)R(S+T)\|=\Big(\frac{3-\sqrt5}2\Big)^2\simeq 0.146<0.276\simeq \|SRS\|. $$ The spirit of the example is that $R$ is the projection onto the eigenspace of $S+T$ corresponding to its smallest eigenvalue, while $SRS$ cuts $R$ "at an angle" and so it takes a bigger chunk.