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
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inequality
1 Answers
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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 $3$.