Bound from below for the power of the sum of two positive operators

Question

Assume $S_1, S_2: H \to H$ are bounded, positive and self-adjoint operators on a real Hilbert space $H$. Clearly, $$\langle (S_1 + S_2)u, u \rangle = \langle S_1 u, u \rangle + \langle S_2 u , u\rangle \geq \langle S_1 u, u \rangle \quad \forall u \in H.$$ One can show that the statement $$ \langle (S_1 + S_2)^m u, u \rangle \geq \langle S_1^m u, u \rangle \quad \forall u \in H, \quad m = 2, 3, \dots \, $$ need not be true, see here. My question now is whether it is possible to find $0 < C < 1$ independent of $S_1, S_2$ (but possibly dependent on $m$) such that $$ \langle (S_1 + S_2)^m u, u \rangle \geq C \, \langle S_1^m u, u \rangle \quad \forall u \in H, \quad m = 2, 3, \dots \, ?$$ A particular case of interest to me is when $H = \mathbb R^n$ and $S_1, S_2$ are positive semi-definite matrices.

Answer 1

In general, no. Take $S_1=\begin{bmatrix}a^2 & a \\ a & 1\end{bmatrix},S_2=\begin{bmatrix}a^2&-a\\-a&1\end{bmatrix},u=\begin{bmatrix}1\\0\end{bmatrix}$ with $a>0$. Then $\langle(S_1+S_2)^m u,u\rangle=(2a^2)^m$ and $\langle S_1^m u,u\rangle\ge a^2$ for all $m$, so for small $a$ we have a problem.