Prove that $\boldsymbol{v}^T\boldsymbol{A}^{-1}\boldsymbol{v}\boldsymbol{z}^T\boldsymbol{A}\boldsymbol{z} \ge (\boldsymbol{v}^T\boldsymbol{z})^2$

Question

I'm facing the following problem:

Prove that

$$ \boldsymbol{v}^\top\boldsymbol{A}^{-1}\boldsymbol{v}\boldsymbol{z}^\top\boldsymbol{A}\boldsymbol{z} \ge (\boldsymbol{v}^\top\boldsymbol{z})^2 $$

where $\boldsymbol{v},\boldsymbol{z} \in \mathbb{R}^n$ are unit vector ($||\boldsymbol{v}||=||\boldsymbol{z}||=1$) and $\boldsymbol{v}^\top\boldsymbol{z}>0$, $\boldsymbol{A}\in \mathbb{R}^{n \times n}$ is symmetric and positive definite.

I am reasonably sure that the assertion is true because I tested it numerically with a lot of random numbers. But I don't find a way to prove this. There exist any property that I can exploit?

If I am able to prove it in the restricted case $n=3$ and/or $\boldsymbol{A}$ diagonal, it is still fine for me. Obviously, it is nicer to have a more generic proof.

Answer 1

The identity that should be helpful here is the identity

$$\|I\|\leq \|A\|\cdot \|A^{-1}\|$$