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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?

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if $$a>b>0$$ is hold we get $$a(a+b)>a(a-b)$$ and this is true since $$2ab>0$$

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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.

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  • $\begingroup$ Thank you for the fast reply, but if the a on the right side is absolute? $\endgroup$ – nevermind Nov 15 '16 at 17:48
  • $\begingroup$ 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. $\endgroup$ – TravisJ Nov 15 '16 at 18:14
  • $\begingroup$ In a second case is not bigger than 0. I formulated the expression slightly inconveniently. But the RHS would still be absolute. $\endgroup$ – nevermind Nov 16 '16 at 8:16

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