# Inequality of the relation

I am having difficulties showing that:

$\frac{a(a+b)}{a-b} > |a|, \qquad a > b > 0.$

(I am not sure if $a > b$ is necessary, but it holds in cases I am considering).

I assume there should be some simple way to do this, or even a "named" inequality for such a simple relation.

Is there a way to show that it holds for a<0, b>0?

if $$a>b>0$$ is hold we get $$a(a+b)>a(a-b)$$ and this is true since $$2ab>0$$
if $$a<0$$ and $$b>0$$ we get $$a(a+b)<-a(a-b)$$ and this is equivalent to $$2a^2<0$$ which isn't true.
• By "absolute" I think @nevermind means that the hypothesis is that $a>b>0$ so to start with assuming $a<0$ is kind of pointless. – TravisJ Nov 15 '16 at 18:14