Assume $S_1, S_2: H \to H$ are bounded, positive and self-adjoint operators on a real Hilbert space $H$. Clearly, $$\langle (S_1 + S_2)u, u \rangle = \langle S_1 u, u \rangle + \langle S_2 u , u\rangle \geq \langle S_1 u, u \rangle.$$ Is it true that $$ \langle (S_1 + S_2)^m u, u \rangle \geq \langle S_1^m u, u \rangle, \quad m = 2, 3, \dots \, ?$$ For matrices that commute, this is true by diagonalization, but the general case is less clear to me. For $m = 2$ and general matrices, the statement is equivalent to $S_1 S_2 + S_2 S_1$ having only non-negative eigenvalue.
EDIT: I found this question, the accepted answer of which proves that the statement I wrote is not true in the general case.
The statement is not true. Assume $S_1$ and $S_2$ are positive semi-definite matrices and consider $m = 2$. We will prove that there exist $S_1$ and $S_2$ such that the right-hand side of $$(S_1 + S_2)^2 - S_1^2 = S_1 S_2 + S_2 S_2 + S_2^2$$ has a negative eigenvalue. Let $A$ and $B$ be as in the accepted answer here, and let $S_1 = A/s$, $S_2 = s B$ for some $s \in \mathbb R$. As $s \to 0$, the right-hand side above tends to $AB + BA$, which implies that the right-hand side has a negative eigenvalue for small enough $s$.