# Are positive real numbers $x,y$ allowed to be taken out during this proof?

Prove $$\left(x^2 – y^2\right)\left(\frac1y – \frac1x\right) \ge 0$$ where $x$ and $y$ are positive real numbers.

Can I simplify $\frac{(x^2 – y^2)(x-y)}{xy} \ge 0$
and then $(x^2 – y^2)(x-y) \ge 0$ cancelling out the $xy$?

Is this valid because then
\begin{align}
(x+y)(x-y)(x-y)&\ge 0\cdot(x-y)^2\ge 0
\end{align}
which is always true and then prove backwards from here?

#### Solutions Collecting From Web of "Are positive real numbers $x,y$ allowed to be taken out during this proof?"

If $x,y$ are positive, then $xy>0$; so, when you have

$$\frac{(x^2 – y^2)(x-y)}{xy} \ge 0$$

you’re allowed to multiply by $xy$ (and the $\ge$ won’t flip), to get

$$(x^2 – y^2)(x-y) \ge 0$$

And from that indeed you can get

$$(x^2 – y^2)(x-y)=(x+y)(x-y)(x-y)=(x+y)(x-y)^2 \ge 0$$

which is true, since $x+y>0$ and $(x-y)^2\ge 0$.