As the title suggests, I have two operators on Hilbert spaces. They are both unbounded but I have bounded the inverse of $A$ (by showing it is strictly positive.) I seek to bound the composition $A^{-1} B$.
In one of my attempts, I have bounded $A > B > 0$ in the expectation value sense, ie. $\langle A \phi, \phi \rangle > \langle B \phi, \phi \rangle$ for all suitable $\phi$. I'm now hoping for some result of the form $1 > A^{-1}B$ (hopefully in norm). Is such a thing true? Are there any results like that?
For example I have tried somehow using $A + B = A (1 + A^{-1}B)$ and maybe trying some Neumann series tricks, but to no avail.
edit: The operators $A$ and $B$ are symmetric, if that helps.
This already fails in a finite-dimensional inner product space. Counterexample: $$ A=\pmatrix{2&1\\ 1&1}, \ B=\pmatrix{1&1\\ 1&1}, \ C=A^{-1}B=\pmatrix{0&0\\ 1&1}, \ \frac{C+C^T}{2}=\pmatrix{0&\frac12\\ \frac12&1}. $$ We have $1\not\geq C$ (because $I-\frac{C+C^T}{2}$ is indefinite) and $\|C\|_2=\sqrt{2}>1$. In this counterexample, we have $A\ge B\ge0$. So, if we add a sufficiently small but positive $\epsilon I$ to $B$ and add $2\epsilon I$ to $A$, we will get a counterexample with $A>B>0,\ 1\not\gt C$ and $\|C\|_2>1$.