Where is the mistake of this derivation?

Question

It is asked to find which on is greater: $$A) \frac{x^2}{y+\frac{1}{y}}$$ $$B) \frac{y^2}{x+\frac{1}{x}}$$ It is given that, $xy\neq0 $ & $x>y$. So I solve this like this $$x>y$$ $$\implies x^3+x>y^3+y$$ $$\implies \frac{1}{x^3+x}<\frac{1}{y^3+y}$$ $$\implies \frac{x^2 y^2}{x^3+x}<\frac{x^2 y^2}{y^3+y}$$ $$\implies \frac{x^2 y^2}{x^2(x+1/x)}<\frac{x^2 y^2}{y^2(y+1/y)}$$ $$\implies \frac{y^2}{(x+1/x)}<\frac{x^2}{(y+1/y)}$$ But if I use $x=2 $ & $y=-2$ then $$\frac{y^2}{(x+1/x)}=\frac{(-2)^2}{(2+1/2)}=\frac{8}{5}$$ $$\frac{x^2}{(y+1/y)}=\frac{2^2}{(-2+1/(-2))}=-\frac{8}{5}$$ which gives total opposite result. Could you please show me where I did wrong?

Answer 1

$$a>b$$

Doesn't imply

$$\frac1a <\frac 1b$$

This is true only if both $a$ and $b$ are positive.

For example $$3>2 \implies \frac 13<\frac 12$$

So far so good, but $$-2>-3 \not \implies -\frac{1}{2}<-\frac 13$$

And

$$2>-3 \not \implies \frac{1}{2}<-\frac 13$$