Question

Using |a| ≤ b ↔ -b ≤ a ≤ b, for all real number x, prove -|x| ≤ x ≤ |x|?

Answer 1

whats the first step? its pretty easy

Answer 2

no clue :/

Answer 3

I can prove considering cases x = 0, x < 0, and x > 0. But the direction says use |a| ≤ b ↔ -b ≤ a ≤ b.