Calculus - inequality problem. I have this inequality :

$$|g(x)-B|&lt;\frac{|B|}{2}$$ 
$$-\frac{|B|}{2}&lt;g(x)-B&lt;\frac{|B|}{2}$$
$$B-\frac{|B|}{2}&lt;g(x)&lt;B+\frac{|B|}{2}$$

I don't understand how it possible to conclude from the inequality that

$$\frac{|B|}{2}&lt;|g(x)|$$ 

Thanks.
$$|B-\frac{|B|}{2}|&lt;|g(x)|$$

Case 1: $$B\geq0$$
$$|B-\frac{|B|}{2}|=\frac{B}{2}=\frac{|B|}{2}&lt;|g(x)|$$

Case 2: $$B&lt;0$$
$$|B-\frac{|B|}{2}|=\frac{-3B}{2}=\frac{3|B|}{2}&lt;|g(x)| \implies \frac{|B|}{2}&lt;|g(x)|$$
If $B&gt;0$ then the first inequality implies $B/2 &lt; g(x)&lt; 3B/2$.
When $B&lt;0$, the first inequality implies $B/2&gt;g(x)&gt;3B/2$