Absolute value equality on $4$ integers

Question

For all $a,b,c,d \in \mathbf{Z},\\a<b<c<d.$

Prove $\left|10-a-b\right|+\left|10-b-c\right|+\left|10-c-d\right|\space = \left|10-a-c\right|+\left|10-a-d\right|+\left|10-b-d\right|$

Is this true? If so, why? I really don't know how to approach this, sorry.

Answer 1

Hint. Clearly, the $10$ can be replaces by any even integer, as $\tilde a=a-5$, $\tilde b=b-5$, $\tilde c=c-5$, $\tilde d=d-5$, also satisfy the same assumptions. So we need to show that

$\left|a+b\right|+\left|b+c\right|+\left|c+d\right|\space = \left|a+c\right|+\left|a+d\right|+\left|b+d\right|$.

The above is clearly NOT always true if $a,b,c,d\ge 0$ or $a,b,c,d\le 0$, unless $$ b+c=a+d. $$

It remain to see what happens when:

i. $a>b>0>c>d$

ii. $a>0>b>c>d$.

Note. Does the exercise say that the $a,b,c,d$ are in arithmetic progress?