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Given that $0\leq \epsilon\leq 1$, $a,b>0$, how to prove $$\frac{1}{(1+\epsilon)^2}\leq \frac{a}{b}\leq (1+\epsilon)^2\implies |a-b|\leq 16\epsilon b?$$

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Welcome to math.SE. Given that you're new here, I'll try to give you some advice: 1. If your question is homework, tag it as such. This will hint the users that you need guidance or hints, rather than a full solution. 2. It is good to include your work on the problem and where it comes from, since it will hint users on which tools to use when solving the problem. – Pedro Tamaroff Apr 10 '12 at 21:50

wlog assume $a \ge b$ (Why?)

Then we have

$$1 \le \frac{a}{b} \le 1 + 2\epsilon + \epsilon^2$$


$$0 \le \frac{a}{b} - 1\le 2\epsilon + \epsilon^2$$

For $0 \le \epsilon \le 1$, it is easy to see that $2\epsilon + \epsilon^2 \le 16 \epsilon$

In fact, you can replace $16$ by $3$.

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