Proof involving a simple inequality

Question

I'm struggling to prove the following inequality (for $0 \leq a \leq c$ and $0 \leq b \leq c$) $$|a-b| \leq c.$$ I see that it is true by drawing a number line and putting the numbers $a, b$ and $c$ there, but I have no idea how to prove it formally. It seems so trivial that it's tempting to write "obviously the following holds next to it." I tried playing around with triangle inequality but I got only this: $$|a-b| \leq |a-c| + |b-c|=2c-a-b,$$ which is not helpful at all.

score 2 · Accepted Answer · answered Mar 15 '21 at 18:05

2

$$-c \leq -b \leq a-b \leq a \leq c$$ hence $|a-b|\leq c$.

answered Mar 15 '21 at 18:05

hgmath

1,830

score 2 · Answer 2 · answered Mar 15 '21 at 18:05

2

$-c=0-c\leq a-b\leq c-0=c$, when you use inequalities that are given for $a$ and $b$.

answered Mar 15 '21 at 18:05

alans

6,475

score 2 · Answer 3 · answered Mar 15 '21 at 18:17

Just to give a different approach, note that

$$|a-b|=\max(a,b)-\min(a,b)\le\max(a,b,c)-\min(0,a,b)$$

holds for any three real numbers $a$, $b$, and $c$. (That is, adding extra options can only make max's bigger and min's smaller.) The assumptions $0\le a,b\le c$, so that $\max(a,b,c)=c$ and $\min(0,a,b)=0$, now do the rest.

score 2 · Answer 4 · answered Mar 15 '21 at 18:37

2

What is $\color {red}{\text{max} |a-b| }$?

$$\begin{align}\color {gold}{\boxed {\color{black}{\text{max} |a-b| =|c-0|=|0-c|=c≤c.}}}\end{align}$$

answered Mar 15 '21 at 18:37

lone student

14,709

Proof involving a simple inequality

4 Answers4