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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.

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Notice that by the famous triangle inequality $(|a+b| \leq |a| + |b|)$

$$ |b| = |b + b_n - b_n| \leq |b - b_n| + |b_n| < \frac{ |b| }{2} + |b_n| $$

Thus,

$$ |b_n| > |b| - \frac{|b|}{2} = \frac{ |b| }{2} $$

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...

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