I've been stuck on this problem for a while, not sure how to tackle it.

$$\text{Let}\ a,b \in \mathbb{R}.\ \text{If}\ 0 < \epsilon < \text{min}\{|a|,|b|\}\ \text{show that}$$ $$\left \vert \frac{a + \epsilon}{b + \epsilon} \right \vert \leq \frac{|a| + \epsilon}{|b| - \epsilon}$$

You're just meant to apply the triangle inequality but I'm not sure how to split up the LHS or how to get the negative on the RHS. Also not sure where to use $e < min\{|a|,|b|\}$ :/

Any help is appreciated.

  • 2
    $\begingroup$ Hint: note that $\left|\frac xy\right| = \frac{|x|}{|y|}$, and then note that each denominator is positive so you can cross multiply and get an equivalent inequality. $\endgroup$ – Mees de Vries Sep 10 '18 at 12:44

Fact 1: If $a,b$ are real numbers and $a<b$ and $c>0$ then $ac<bc$

Now by triangle inequality

$|a+\epsilon|\leq |a|+|\epsilon|=|a|+\epsilon$

as $\epsilon$ is positive number.

So, using fact $1$,

$$\left \vert \frac{a + \epsilon}{b + \epsilon} \right \vert \leq \frac{|a| + \epsilon}{|b+ \epsilon \vert}\tag1\label {eq1}$$

Now using the reverse triangle inequality $|a+b|\geq||a|-|b||$

$|b+\epsilon|\geq ||b|-\epsilon|=|b|-\epsilon$

This follows from the condition given on $\epsilon$

So $$\frac{1}{|b+\epsilon|}\leq \frac{1}{|b|-\epsilon}$$

Again using the fact 1,

$$\frac{|a| + \epsilon}{|b+ \epsilon \vert}\leq \frac{|a| + \epsilon}{|b|-\epsilon}\tag2 $$

Using $(1)$ and $(2)$, you get the desired inequality.

  • 1
    $\begingroup$ Thanks so much, didn't think of doing it like this $\endgroup$ – tzcl Sep 11 '18 at 1:07

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