$ Ax\cdot x>0$ and $Ay\cdot y>0$ implies $(Ax\cdot x)(Ay\cdot y)\geq (Ax\cdot y)^2$?

Question

Let $A$ be a $n\times n$ symmetric real matrix. Assume that $x,y\in\mathbb{R}^n$ are such that $Ax\cdot x>0$ and $Ay\cdot y>0.$ Does this imply that $(Ax\cdot x)(Ay\cdot y)\geq (Ax\cdot y)^2$? The inequality clearly looks like a Cauchy-Schwarz (C-S) type inequality but applying C-S I couldn't arrive to the desired inequality. Besides, I am not even sure if it is even true. Does anyone have any thoughts?

Answer 1

For a counterexample, suppose we have $$A = \begin{bmatrix} 1 & 5 \\ 5 & 1 \end{bmatrix}, x = e_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, y = e_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}.$$ Then $Ax \cdot x = 1$ and $Ay \cdot y = 1$, but on the other hand, $Ax \cdot y = 5$.

On the other hand, the inequality is true if $A$ is a positive definite symmetric matrix (or even positive semidefinite). The idea here is: if $A$ is positive definite symmetric, then that implies $\langle x, y \rangle := Ax \cdot y$ forms an inner product on $\mathbb{R}^n$, and the Cauchy-Schwarz inequality for this inner product gives exactly the desired result.

Answer 2

If $A$ is postive definite, then has a square root, e.g., $\sqrt{A}$ exists.

Consider the inner product \begin{align} \langle\sqrt{A}x,\sqrt{A}y\rangle = x^TAy \end{align}

Assuming this is a valid inner product (I didn't verify, but looks like it should be), your result follows by direct application of the CS inequality.