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?



Inverse quadratic norms
The proposed result holds **true**.


I am assuming throughout that $spd$ means [*symmetric positive definite*](https://en.wikipedia.org/wiki/Definite_matrix) 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](https://en.wikipedia.org/wiki/Symmetric_bilinear_form)), 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](https://en.wikipedia.org/wiki/Definite_matrix) 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.


