Proof $|a-b| \lt \frac{|a|}{2} \rightarrow |b| \gt \frac{|a|}{2}$

Question

In this answer to a question of mine, in the second paragraph, the author uses an expression analogous to $|a-b| \lt \frac{|a|}{2}$ , and then in the parenteses says that it implies that $|b| \gt \frac{|a|}{2}$.

It didn't seem obvious to me so I tried to prove it:

I assumed $|a-b| \lt \frac{|a|}{2}$, which is equivalent to $-\frac{|a|}{2} \lt a-b \lt \frac{|a|}{2}$. Adding b to both sides gets

$$ b-\frac{|a|}{2} \lt a \lt b+ \frac{|a|}{2}$$

Noting that $b \leq |b|$ and $-|b| \leq b$ we obtain

$$-|b| - \frac{|a|}{2} \lt a \lt |b| + \frac{|a|}{2}$$

$$\iff |a| \lt |b| + \frac{|a|}{2} \iff |b| \gt \frac{|a|}{2}. \square $$

This was not too easy for me so my questions are:

$\bullet$ Is this proof correct?

$\bullet$ Is there an easier way to prove this?

$\bullet$ How is a more experienced mathematician able to deal with these inequalities so easily, that (s)he doesn't need to write them down when using them in a proof?

Answer 1

Your proof is correct.

You can also use triangle inequality: $$|a| \le |a-b|+|b| < \frac{|a|}{2}+|b| \implies \frac{|a|}{2} < |b|.$$

Answer 2

Remember that $|x|^2 = x^2$. If you square it you get $$4(a-b)^2< a^2$$ so $$(2a-2b-a)(2a-2b+a)<0$$ or $$(a-2b)(3a-2b)<0$$

solving this quadratic inequality on $b$ we conclude that $$\boxed{\color{red}{{a\over 2}<b<{3a\over 2}}}$$

Answer 3

When I read your hypothesis, $|a-b| \lt \frac{|a|}{2}$, this is what I mentally translated it into: The distance from $a$ to $b$ is less than half the distance from $a$ to $0$. I concluded that $b$ must be closer to $a$ than to $0$; that is, $b$ is further from $0$ (in the direction toward $a$) than the halfway point $\frac{a}{2}$ is. In symbols, that becomes $|b| \gt |\frac{a}{2}|$.