Prove that $|a|\leq \max\{|b|,|c|\}$ if $b\leq a \leq c$

Question

Prove that $|a| \leq \max\{|b|,|c|\}$ if $b\leq a \leq c$.

I showed that $a\leq c$ and therefore $-c\leq a \leq c$ so that $|a|\leq c$ but then I got stuck.
Is this the right approach?

Answer 1

• $|a| = a \text{ or } -a$
• $a \leq c \leq |c|$
• $- a \leq -b \leq |b|$

Answer 2

Hint:

Simply break the problem into 3 cases.

Case 1: $b < 0 \leq c.$

Case 2: $b < c < 0.$

Case 3: $0 \leq b < c.$

Then manually attack each case.

Answer 3

Alternative proof: Consider the function $y=x^2$ where $x\in [b,c]$. By Fermat's theorem (or just property of parabolas) there is no local maximum. Therefore $$a^2 \leqslant \max(b^2,c^2) \implies |a| \leqslant \max(|b|, |c|).$$