Proving inequality using AM-GM inequality

Question

Question:

$ Suppose\ x,\ y\ are\ positive\ real\ numbers.\ Show\ that\ \left(x^2-y^2\ \right)\left(\frac{1}{y}-\frac{1}{x}\right)\ge 0 $

My attempt:

I tried proving it using AM-GM inequality. Is my approach correct? If not, how can I prove it using AM-GM inequality?

Answer 1

It's just $$\frac{(x-y)^2(x+y)}{xy}\geq0$$

Answer 2

If $x = y$, the inequality holds.

If $y > x$, then $x^2 < y^2$ (hence $x^2 - y^2 < 0$) and $1/y < 1/x $ (hence $1/y - 1/x < 0$) and the inequality holds.

Do the same for $y < x$.