inequalty concering inverses of positive definite matrix

Question

I don't find a way to prove this: given $A$, $B$, symmetric and positive definite:

$$A>B \Rightarrow A^{-1} < B^{-1},$$ where $A>B$ means that $A-B$ is positive definite.

Answer 1

First, assume we have solved it when $A=I$. We have, as $A>0$, that $A$ admit a positive define square root $A^{1/2}$. We have
$I-A^{-1/2}BA^{-1/2}>0$. Let $B':=A^{-1/2}BA^{-1/2}>0$. Then $B'^{—1}>I$, hence $A^{1/2}B^{-1}A^{1/2}>I$ and we are done.

Now we solve this case: write $B:=C^2$, where $C>0$. Then for $x\neq 0$, $\lVert Cx\rVert^2<\lVert x\rVert^2$, which gives $\lVert y\rVert^2<\lVert C^{—1}y\rVert^2$ for $y\neq 0$. This gives $C^{—2}>I$ hence $B^{-1}>I$.