$|a-b|+|b-c|+|c-a|=2(\max\{a,b,c\}-\min\{a,b,c\})$ 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
 A: $1$. 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 $2(c-a)$.
A: 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.
A: Use these following equivalences (how to deduce this equality) 
$$\max(a, b) = \dfrac{a + b + |a - b|}{2},$$
$$\min(a, b) = \dfrac{a + b - |a - b|}{2},$$
then
$$\max(a, b) - \min(a, b) = |a - b| $$
therefore
$$\max(a, b, c) = \max(\max(a,b),c) = \dfrac{\max(a, b) + c + |\max(a,b) - c|}{2}$$
$$\min(a, b, c) = \min(\min(a,b),c) = \dfrac{\min(a, b) + c - |\min(a,b) - c|}{2}$$
concluding
$$ \max(a, b, c) - \min(a, b, c) = $$
$$ = \dfrac{\max(a, b) + c + |\max(a,b) - c|}{2} - \dfrac{\min(a, b) + c - |\min(a,b) - c|}{2} = $$
$$ = \dfrac{\max(a, b) - \min(a, b) + |\max(a,b) - c| + |\min(a,b) - c|}{2} = $$
$$ = \dfrac{|a-b| + |\max(a,b) - c| + |\min(a,b) - c|}{2} = $$
choose $ a > b $ or $ b > a $ without loss of generality
$$ = \dfrac{|a-b| + |a - c| + |b - c|}{2} $$
