Why is it that $\left|b_n - b \right| < \frac{\left|b \right|}{2} \Rightarrow \left| b_n \right| > \frac{\left|b \right|}{2}$? Unfortunately I am stuck on one step of a proof for an algebraic limit theorem, specifically:
Why is it exactly that $\left|b_n - b \right| < \frac{\left|b \right|}{2} \Rightarrow \left| b_n \right| > \frac{\left|b \right|}{2}$ ?
If this doesn't make sense without more context, please let me know. Otherwise, thank you for your help!
 A: There's a version of the triangle inequality that says $\big| \,|x| - |y| \,\big| \leq |x - y|$ for all $x$ and $y$. So you have
$$\big|\,|b| - |b_n|\,\big| \leq |b - b_n| < {|b| \over 2}$$
So in particular you have
$$|b| - |b_n| < {|b| \over 2}$$
Rearranging this expression gives what you want.
A: Hint: use $|x|\le|y|+|x-y|$ hence $|y|\ge |x|-|x-y|$ for suitable values of $x$ and $y$.
A: You may divide everything by $b$, then this is equivalent to the statement:
 $|x-1|\lt 1/2$ implies $x\gt1/2$. 
In other words: if $x$ is at a distance less than $1/2$ from $1$ then $x$ must be greater than $1/2$. 
A: Rewriting the antecedent, we want to show $\frac{|b|}{2} > |b_n-b| \Rightarrow |b_n| > \frac{|b|}{2}$.
\begin{align*}
\frac{|b|}{2} \color{red}{+|b_n| - \frac{|b|}{2}} = \color{LimeGreen}{|b_n|} &> |b_n-b| \color{red}{+|b_n| - \frac{|b|}{2}} \\
&= |b-b_n| \color{red}{+|b_n| - \frac{|b|}{2}} \\
&> |b-b_n+b_n| - \frac{|b|}{2} \\
&= |b| - \frac{|b|}{2} \\
&= \color{LimeGreen}{\frac{|b|}{2}}
\end{align*}
