<s>[INST] Where is the mistake of this derivation? 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? 
 [/INST] $$a>b$$
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
$$2>-3 \not \implies \frac{1}{2}<-\frac 13$$
</s>