# Unsure How to Proceed with Proof Related to Algebraic Limit Theorem

I'm independently studying Stephen Abbott's Understanding Analysis and am trying to follow the proof for

$$(b_n) \rightarrow b$$ implies $$(\frac{1}{b_n}) \rightarrow (\frac{1}{b})$$

Specifically, I'm lost where Abbott says

Consider the particular value $$\epsilon_0 = \frac{|b|}{2}$$. Because $$(b_n) \rightarrow b$$, there exists an $$N_1$$ such that $$|b_n - b| < \frac{|b|}{2}$$ for all $$n \geq N_1$$. This implies $$|b_n| > \frac{|b|}{2}$$.

Why does $$|b_n - b| < \frac{|b|}{2}$$ imply $$|b_n| > \frac{|b|}{2}$$? In the book, the only properties of the absolute value function that have been covered are the following.

\begin{align} |ab| &= |a||b| \\ |a+b| &\leq |a| + |b| \end{align}

I guess I could solve this question by showing that this holds for every possible "+/- combination" of $$b_n$$ and $$b$$, but I was wondering if there was a cleaner way to show this.

Notice that by the famous triangle inequality $$(|a+b| \leq |a| + |b|)$$
$$|b| = |b + b_n - b_n| \leq |b - b_n| + |b_n| < \frac{ |b| }{2} + |b_n|$$
$$|b_n| > |b| - \frac{|b|}{2} = \frac{ |b| }{2}$$
Added: Rephrase of the paragraph: Because $$(b_n) \to b$$, the definition of the limit says that $${\bf for \; any}$$ $$\varepsilon > 0$$ ( for example, take $$\varepsilon = \frac{ |b| }{2} > 0$$), then $$|b_n - b| < \frac{|b|}{2}$$ and then the rest follows...