$|a-b|+|b-c|+|c-a|=2(\max\{a,b,c\}-\min\{a,b,c\})$

Question

Let $a,b,c ∈ \Bbb R$ Show that

$|a-b|+|b-c|+|c-a|=2(\max\{a,b,c\}-\min\{a,b,c\})$

Not sure where to start

Answer 1

Start by assuming that $a\le b\le c$. Then $${\rm LHS}=(b-a)+(c-b)+(c-a)=2c-2a={\rm RHS}\ .$$ There are five other cases to consider but if you think carefully you might find a short cut. Good luck!

Edit: for the short cut, see @Martin's answer.

Answer 2

$1$. Step: Show that both sides of your equation are invariant unter permutations of $a,b,c$.

Hence, we may assume without loss of generality $a \leq b \leq c$.

$2$. Step: Show that if $a \leq b \leq c$, both sides equal $2(c-a)$.