Absolute Value Inequality Including Itself

Question

Given a real number $a$. Will it be correct to use the following inequality in the proof:

$$-a\le|a|\le a$$

Although "less" and "greater" parts never actually happen, the whole equation will always be valid since depending on the value of $a$ either right or left equality will be true. Therefore the whole inequality is always true.

Is it legit to use this logic in the proof?

Answer 1

In general, it's fine (sometimes desired even) to use less information than you actually have for a proof.

However, for your example, you need to specify $a\ge 0$ to make it valid.

Answer 2

Your equation is absolutely correct and is a commonly used technique (losing the knowledge of equality but setting you up to get a better solution through inequalities). However, your equation only holds for $a \geq 0$, but is otherwise fine. If $a < 0$ then you have $-a \geq |a| \geq a$

Answer 3

Well, you could say $ (min-a, a) \le |a| \le max(-a, a)$ but then it's kind of pointless as $|a| = max(-a, a)$

Basically it's best to simply say |a| = {a whenever a $\ge$ 0; -a whenever a $\lt$ 0);