Can I proove equality of x and y form this expression?

Question

$x$ and $y$ are from $S=(0,1)$, $S \subset \mathbb{R}$. I'm trying to prove that a relation on that set defined as:

$$x\rho y \Leftrightarrow \frac{x^2}{1-y^2} \ge \frac{y^2}{1-x^2}$$

is antisymetrical, but get an expression like this:

$$x\rho y \: \wedge \: y \rho x \Leftrightarrow \frac{x^2}{1-y^2} \: \ge\frac{y^2}{1-x^2} \: \wedge \frac{y^2}{1-x^2} \ge \frac{x^2}{1-y^2} $$

$$\Rightarrow \frac{x^2}{1-y^2} \ge \frac{x^2}{1-y^2}$$

From that I can conclude that the expression above is true for any $x,y \in S$, but not that $x=y$.

But to prove that it's not true I have to provide one example for which it's not true, which I can't do without selecting two same numbers.

How can I prove that it is, or is not antisymetrical?

Answer 1

For $x=\sqrt{1/3}$ and $y=\sqrt{2/3}$ we have $x\rho y \: \wedge \: y \rho x$

Answer 2

Note that you can actually prove $xRy \land yRx \iff x^2(1-x^2) = y^2(1-y^2)$ since $ x, y \notin \left\{ -1,1\right\}$

Now take $y = \sqrt{1-x^2} \ne x \forall x \ne \frac{ \sqrt{2}} 2$