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

Then we have

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

i.e

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

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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