# How do I prove that $|c| ≤ \max\{b, −a\}$ given that $a, b, c \in \mathbb R$ and $a ≤ c ≤ b$?

EDIT: I forgot to mention the second condition: $a \leq c \leq b$.

I'm taking a first-year undergraduate calculus course and I'm faced with this problem. I'm not sure about how to proceed without using examples. I realize that $|c|$ might be greater than $b$ or $-a$ depending on the sign of each term.

• You should add some context, as the claim is false as stated. Consider $a=b=0$ and $c=1$. – 211792 Sep 8 '16 at 18:38
• Is there some known relationships between $a, b$ and $c$? – the_candyman Sep 8 '16 at 18:38
• As you have it, the $c$ is not related to $a$ or $b$ in any way, so there is no reason why the inequality should hold in general. For example, take $c=10, b = 1, a = -2$. Is something missing in your problem statement? – avs Sep 8 '16 at 18:39
• @avs Right idea, but incorrect counterexample. – 211792 Sep 8 '16 at 18:40
• Thanks, @AustinC. Corrected. – avs Sep 8 '16 at 18:43

If $c \geq 0$:
$$|c| \leq b \leq \max(b, -a)$$
If $c < 0$:
$$|c| \leq -a \leq \max(b, -a)$$