Q:

Paradox inequality

By doing exercises, I applied a reasoning in a demonstration, that looking at it well, it seems paradoxical.
Let's take a real number $\ a>0 $.
Then $b( \epsilon ) = a+ \epsilon > a $, for each $\ 0< \epsilon <1 $.
$\ b^2( \epsilon ) = (a+ \epsilon )^2 = (a^2 + 2 a \epsilon +\epsilon ^2) < a^2 + 2 a \epsilon + \epsilon $
$\ b^2( \epsilon ) < a^2 + \epsilon (2a+1)$
Since $\ \epsilon $ is as small as I want and it must always be that $\ b^2( \epsilon ) < a^2 + \epsilon (2a+1)$ 
then it means that:
$\ b^2( \epsilon ) \leq a^2 $
so $\ b( \epsilon ) \leq a $, that contradicts $\ b( \epsilon )> a $.
where am I wrong?
Thanks.

A:

Why do you make it so complicated? Here is a much simpler way to get a contradiction using your idea. Define
$$f(\epsilon)=\epsilon\text{ for }\epsilon\gt0.$$ So $$f(\epsilon)\gt0.$$ On the other hand, $$f(\epsilon)\le\epsilon.$$ Since I can make $\epsilon$ as small as I want this means that $$f(\epsilon)\le0,$$ which contradicts $f(\epsilon)\gt0.$

If you can see what I did wrong, maybe you can see that you made the same mistake in your argument.

