If $|a-b| \leq \frac{\epsilon}{2}$ and $|a| \gt \epsilon$, constructively prove that $|b|\geq \frac{\epsilon}{2}$. I have a proof for the claim, but it's ugly and involves cases and a contradiction. Is there a constructive proof for the statement:

If $|a-b| \leq \frac{\epsilon}{2}$ and $|a| \gt \epsilon$, then $|b|\geq \frac{\epsilon}{2}$

I'm guessing it involves the triangle inequality theorem, but I am struggling to pin it down.
Thank you!
 A: Using
$$ |x-y|\ge|x|-|y| $$
one has
$$ |b|=|a-(a-b)|\ge|a|-|a-b|>\epsilon-\frac{\epsilon}{2}=\frac{\epsilon}{2}. $$
A: Hint
$$|a-b|\le\frac{\epsilon}{2}\Leftrightarrow -\frac{\epsilon}{2}\le a-b\le \frac{\epsilon}{2}\Leftrightarrow a-\frac{\epsilon}{2}\le b \le a+\frac{\epsilon}{2}\qquad (1)$$
$$|a|>\epsilon \Leftrightarrow a
<-\epsilon \text{ or } a>\epsilon \Leftrightarrow \left(a+\frac{\epsilon}{2}
<-\frac{\epsilon}{2}\right) \text{ or } \left(a-\frac{\epsilon}{2}>\frac{\epsilon}{2}\right)\qquad(2)$$
Now, put both inequalities, $(1)$ and $(2)$, together. Can you finish?
A: You have to use the inverse triangle inequality, that is for any real numbers $a$ and $b$ we have
$$|a|-|b| \leq |a-b|.$$
On the one hand,
$$|a|-|b| \leq |a-b|\leq \frac{\varepsilon}{2}.$$
Hence,
$$|a|\leq |b|+\frac{\varepsilon}{2}.$$
Also $|a|>\varepsilon$, so
$$\varepsilon<|a|\leq |b|+\frac{\varepsilon}{2}.$$
In particular,
$$\varepsilon<|b|+\frac{\varepsilon}{2},$$
where the inequality is strict because of the fact that $\varepsilon<|a|$, and the result is proven:
$$|b|>\frac{\varepsilon}{2}.$$
