Property of a matrix - $x^T A y\geq x^TAx$ implies $y^TAy>y^TAx$.

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There is a $n\times n$ matrix $A$.

$A$ satisfies the following property for $x,y\in \mathbb R^n$ and $x\neq y$,

$$x^T A y\geq x^TAx\Rightarrow y^TAy>y^TAx.$$

Is there any known matrix property that satisfies the condition?

For symmetric positive definite $A$'s, it is true, but this is a sufficient condition, not a necessary condition. For instance, if $x=(-1,-1)$, $y=(-2,-4)$ and $A=(7,0;0,-1)$, the condition is satisfied but $A$ is not a PD matrix.

Any type of help (hint, literature, solution) will be highly appreciated.

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If a real $n\times n$ matrix $A$ satisfies $$x^T A y\geq x^TAx\implies y^TAy>y^TAx$$ for all $x,y\in \mathbb R^n$ with $x\neq y$ then $A$ is positive definite: For $y \ne 0 = x$ is $$ x^T A y = 0 = x^TAx\implies y^TAy>y^TAx = 0 \, . $$

Conversely, if $A$ is positive definite and $x \ne y$ with $x^T A y\geq x^TAx$ then $$ y^TAy - y^TAx = \underbrace{(x-y)^T A (x-y)}_{> 0} + \underbrace{(x^T A y - x^T A x)}_{\ge 0} > 0 \,. $$