<s>[INST] Proving the equality case in triangle inequality Background
When plotted on a real number line, it may be deduced that if 
$$a,b,c \in \mathbb{R} $$
$$a < b < c$$
then 
$$\left| {a - c} \right| = \left| {a - b} \right| + \left| {b - c} \right|$$
Problem
But the problem is with the proof. How can the above statement be proven true from the properties of order structure and the definition of absolute value? 
 [/INST] To prove it you do cases
$|a-c| = \pm (a-c)$ and $|a-b| + |b- c| = \pm(a-b) + \pm (b-c)$.
consider the ways that $a,b,c$ may be ordered:
(your case is a subcase of case 6: so you can skip to the very last case if you want.)
1:  Assume $a \ge b \ge c$ then
$|a-c| = a-c$ and $|a-b| + |b-c| = (a-b) + (b-c) = a-c = |a-c|$.
2:  Assume $a \ge c \ge b$ then 
$|a-c| = a-c$ and $|a-b| + |b-c| = (a-b) -(b-c) = a-2b + c$.  Now $2b \le 2c$ so $-2b \ge -2c$ and $a-2b + c\ge a -2c +c = a-c = |a-c|$.
3: Assume $b \ge a \ge c$ then
$|a-c| = a-c$ and $|a-b| + |b-c| = -(a-b) +(b-c) = -a+2b - c$.  Now $2b \ge 2a$ so  $-a+2b - c \ge  -a +2a -c = a-c = |a-c|$.
4: Assume $b \ge c \ge a$
$|a-c| = -(a-c) = c-a$ and $|a-b| + |b-c| = -(a-b) +(b-c) = -a+2b - c$.  Now $2b \ge 2c$ so  $-a+2b - c \ge  -a +2c -c = -a+c = |a-c|$.
5: Assume $c \ge a \ge b$
$|a-c| = -(a-c) = c-a$ and $|a-b| + |b-c| = (a-b) -(b-c) = a-2b + c$.  Now $2b \le 2a$ so $-2b \ge -2a$ and $a-2b + c\ge a -2a +c = -a+c = |a-c|$.
6: Assume $c \ge b \ge a$
$|a-c| = -(a-c) = c-a$ and $|a-b| + |b-c| = -(a-b) - (b-c) = (b-a) + (c-b) = -a + c = |a-c|$.
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The "equality cases are 1) and 6) and they are very straightforward.
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