# Inequality of operators $\left\vert \langle x, TS x \rangle \right\rvert \le \langle x,T x \rangle \left\lVert S \right\rVert$

Let $T$ be a bounded symmetric positive operator on a complex Hilbert space $H$ and $S$ a bounded operator.

Is it then true that for all $x \in H$

$$\left\vert \langle x, TS x \rangle \right\rvert \le \langle x,T x \rangle \left\lVert S \right\rVert?$$ Looks like something that could be true, but I did not manage to conclude it from simple algebra.

No, even in dimension $2$.
Let $S(x,y) = (x+y,0)$; then $\|S\| = \sqrt{2}$. Let $T(x,y) = (x,0)$. We have $$\langle (x,y), T S (x,y) \rangle = \bar{x} (x + y)$$ and $$\langle (x,y), T (x,y) \rangle \|S\| = \sqrt{2} |x|^2 ,$$ so the inequality does not hold, for example, for $(x,y) = (1,1)$.