How do I prove that $\left|b_n-b\right|<\frac{|b|}{2} $ implies $\left|b_n\right|>\frac{|b|}{2}$

Question

It's a super important step in the Abbott Analysis book that:

$\left|b_n-b\right|<\frac{|b|}{2} $

implies

$\left|b_n\right|>\frac{|b|}{2}$

And I can't for the life of me figure how to prove it even though it's a simple step :( Tried the triangle inequality in 50 different configurations

When I draw it out with all the cases I get it, but I want to be able to prove it without considering all the positives and negative combinations separately

Answer 1

$$|b|\leq |b-b_{n}|+|b_{n}|<\frac{|b|}{2}+|b_{n}| \implies \frac{|b|}{2}<|b_{n}|$$