If $a\le b\le c$, then $|b-d|\le\max\{|a-d|,|c-d|\}$

Question

I am trying to find a simple way to show the fact that

if $a\le b\le c$, then $|b-d|\le\max\{|a-d|,|c-d|\}$ for any number $d$.

Is there a way to do this besides breaking it up into the cases

1) $d\le a\;\;\;$ 2) $a\le d\le c\;\;\;$ 3) $d\ge c$?

Answer 1

"Simple" is a subjective term, but I think this way is simpler as we need only check two cases:

Let $a\leq b\leq c$, now $a-d\leq b-d\leq c-d$.
If $b-d\geq 0$, then $|b-d|\leq |c-d|\leq \max\{|a-d|,|c-d|\}$.
If $b-d\leq 0$, then $|b-d|\leq |a-d|\leq \max\{|a-d|,|c-d|\}$.
Therefore $|b-d|\leq \max\{|a-d|,|c-d|\}$.

Answer 2

The numbers $|b-d|$, $|a-d|$, and $|c-d|$ are simply the distances from $d$ to the points $b,a$, and $c$, respectively. If $d<b$, then $d$ is closer to $b$ than to $c$, so $|b-d|<|c-d|\le\max\{|a-d|,|c-d|\}$. If $d>b$, then $d$ is closer to $b$ than to $a$, so $|b-d|<|a-d|\le\max\{|a-d|,|c-d|\}$. And the case $b=d$ is trivial.