Given that $0\leq \epsilon\leq 1$, $a,b>0$, how to prove $$\frac{1}{(1+\epsilon)^2}\leq \frac{a}{b}\leq (1+\epsilon)^2\implies |a-b|\leq 16\epsilon b?$$
Proof of an inequality in the Change of Variables formula proof
2
$\begingroup$
inequality
1 Answers
2
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 $3$.