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Re: is x^2 < x - |y|?

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WholeLottaLove wrote:
Ok, I think I understand. We have two cases for |y|...y and -y. In the case of x^2 < y... it is derived from the inequality x^2 < x -(-y) right?


The passages above were correct, so I think you got the point.

Yes, from x^2<x+y you sum x<0 and obtain x^2<y, that's correct

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