Show that $|a-b|<\frac{|b|}{2}$ implies $\frac{|b|}{2}<|a|$ Pretty simple, for $a,b \in \mathbb R$, show that $|a-b|&lt;\frac{|b|}{2}$ implies $\frac{|b|}{2}&lt;|a|$.  I can see this graphically on the number line, but I can't seem to show it algebraically.
I'm think it involves the triangle inequalities:

$$|x+y|\le|x|+|y|$$
$$|x-y|\ge||x|-|y||.$$
First, rewrite $|a-b|&lt;\cfrac{|b|}{2}$ as $|b-a|&lt;\cfrac{|b|}{2}$. 

Now, I'll use your second version of the triangle inequality (without the extra bars) and we obtain
$$|b|-|a| \leq |b-a|&lt;\frac{|b|}{2}$$
which means
$$|b|-|a| &lt; \frac{|b|}{2}$$
Finally, add $|a|$ and substract $|b|/2$ to both sides of the latter to get the desired result.
$$ |a-b|&lt;\frac {|b|}{2} \implies -\frac {|b|}{2} &lt; a-b&lt;\frac {|b|}{2}$$

$$\implies b-\frac {|b|}{2} &lt;a&lt; b+\frac {|b|}{2}$$

For $b&gt;0$ we get $$\frac {b}{2} &lt;a&lt;\frac {3b}{2}\implies \frac {b}{2} &lt;a $$
for $b&lt;0$  we get $$\frac {3b}{2} &lt;a&lt;\frac {b}{2}\implies -\frac {b}{2} &lt;- a$$
 Thus  in any  case  we have $$\frac {|b|}{2}&lt; |a|$$