# Proving inequality using AM-GM inequality

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?

• The square of any real number is non-negative, so I don't see why you need to prove that $(x-y)^2 \ge 0\,$. – dxiv May 10 '17 at 20:16
• So the work you went through using AM-GM is actually how the inequality is typically proved. $(x-y)^2$ is always a positive quantity, so you can just use this fact without appealing to AM-GM – helloworld112358 May 10 '17 at 20:17
• I understand that it is obvious but I how do I formally explain my steps? – user444945 May 10 '17 at 20:18
• At the level where you know and are allowed to use AM-GM, I presume you don't need a proof that $z^2 \ge 0\,$ for all $\,\forall z \in \mathbb{R}\,$. – dxiv May 10 '17 at 20:20
• What about (x + y) / xy – user444945 May 10 '17 at 20:21

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

• That's what the OP wrote, but tried to prove $(x-y)^2 \ge 0$ with AM-GM. – dxiv May 10 '17 at 20:18
• @dxiv I did not see it. Now I see that AM-GM here it's not necessary. – Michael Rozenberg May 10 '17 at 20:21
• Could you start from the beginning and explain your steps? Thanks. – user444945 May 10 '17 at 20:22
• @Josh Mitkitzel Now I see that you got it in your seventh line. – Michael Rozenberg May 10 '17 at 20:25
• Do you think thats a good enough explanation? – user444945 May 10 '17 at 20:26

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 $y < x$.

• I don't quite understand. Could you show me some steps? – user444945 May 10 '17 at 20:30
• @JoshMitkitzel You have a product of the numbers $x^2 - y^2$ and $1/y - 1/x$, and you must show that the product is not smaller than $0$. This will be true if both terms are $> 0$ or both terms are $<0$ or at least one of the terms $= 0$. – Antoine May 10 '17 at 20:32