Suppose that for a given fixed $v$, and two positive definite symmetric matrices $A$ and $B$ (with real-valued entries) it holds that: $$ v^T A v > v^T B v. $$ Does it then hold that $$ v^T A^{-1} v < v^T B^{-1} v $$ for the same $v$?
Context: I am working with some inequalities involving quadratic forms of the form $v^TA^{-1}v$ where $A$ is the expected value of a matrix-valued random variable. I can bound the expectation $v^TAv$ both above and below but not its inverse so easily, so if I could get rid of the inversion it would be helpful, but I'm not convinced that I can. Of course there may also be another way to approach problem.
What I have tried: I considered some basic approaches using spectral decompositions of $A$ and $B$ but since the eigenvectors can be different I didn't get very far. I am not, however, very experienced in this area. I also tried to find a simple counterexample among 2 x 2 matrices. I couldn't but that does not mean that there isn't one!
This is not correct. Try $n=2$ and $A=I_2$ and $B$ diagonal.