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

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?

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.
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}