Proving the equality case in triangle inequality

Question

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?

Answer 1

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 $|a-b| + |b-c| = -(a-b) - (b-c) = (b-a) + (c-b) = -a + c = |a-c|$.

===

The "equality cases are 1) and 6) and they are very straightforward.

Answer 2

Since $a<b<c$ we have

$|a-c|=c-a, |a-b|=b-a$ and $|b-c|=c-b$.

This gives:

$$\left| {a - b} \right| + \left| {b - c} \right|=b-a+c-b=c-a=|a-c|.$$

Answer 3

Recall that $$|x| = \left\{\begin{matrix}+x & \text{if } x\ge 0;\\-x & \text{otherwise}.\end{matrix}\right.$$ Using the above-mentioned definition and $a < b < c$, we get $|a-b|=b-a$, $|b-c|=c-b$, and $|a-c|=c-a$. Consequently, .....

Answer 4

Alternatively, equality in the triangle inequality $$|x+y|\leq |x|+|y|$$ holds if and only if $x$ and $y$ have the same sign; that is if and only if $x\cdot y\geq0$.

Therefore, equality holds in $$|x-z|\leq |x-y|+|y-z|$$ if and only if $$(x-y)(y-z)\geq 0\,.$$ This is quite easy to show if $x<y<z$.

Answer 5

Since both sides of the equality are nonnegative, we can square it:

\begin{align} |a-c| = |a-b| + |b-c| &\iff |a-c|^2 = (|a-b| + |b-c|)^2\\ &\iff a^2-2ac+c^2 = a^2-2ab+b^2+b^2-2bc+c^2 + 2|a-b||b-c|\\ &\iff 0 = b^2 - ab - bc + ac + |a-b||b-c|\\ &\iff 0 = -(a-b)(b-c) + |a-b||b-c|\\ &\iff (a-b)(b-c) = |a-b||b-c|\\ &\iff a-b \,\text{ and } \,b-c \text{ have the same sign}\\ &\iff a \ge b \ge c \,\text{ or }\,a \le b \le c \end{align}