The following problem seems easy at a first glance but I can't see the way to prove it. Actually I don't even know if it's true but it is assumed implicitly in a research paper.
Help highly appreciated.
Given two spd matrices $A$, $B$ with
$x^\top Ax \ge x^\top Bx$
then it follows that $x^\top A^{-1}x \le x^\top B^{-1}x$.
Any ideas?
The proposed result holds true.
I am assuming throughout that $spd$ means symmetric positive definite and that the matrices $A$ and $B$ are $n$-by-$n$ matrices over $\mathbb{R}$ for some $n > 0$.
Indeed, since $B = P^{\top}P$ for some $P \in GL_n(\mathbb{R})$ by hypothesis (see e.g., Sylvester's law of inertia), we can assume, without loss of generality, that $B = I_n$, the $n$-by-$n$ identity matrix.
Let us assume that $A - B = A - I_n$ is positive semi-definite and let $O$ be an $n$-by-$n$ orthogonal matrix over $\mathbb{R}$ such that $O^{\top}AO$ is diagonal. Clearly, the matrices $O^{\top}(A - I_n)O$ and $O^{\top}(I_n - A ^{-1})O$ are also diagonal. Conjugating then both sides of the identity $A - I_n = A(I_n - A^{-1})$ by $O$, yields immediately that $I_n - A^{-1}$ is positive semi-definite (check the signs of diagonal coefficients on both sides), hence the result.
The result follows from operator monotonicity of inverse over symmetric positive definite matrices, i.e., $A \succeq B \Rightarrow A^{-1} \preceq B^{-1}$, which is a special case of the Löwner-Heinz theorem. See for example, Theorem 2.6 here.
