The Triangle Inequality

Theorem

For all numbers $a$ and $b$ , we have

|a+b| \leq |a| + |b|

We will consider 4 cases:

(1) \quad a \geq 0,\quad b \geq 0

(2) \quad a \geq 0,\quad b \leq 0

(3) \quad a \leq 0,\quad b \geq 0

(4) \quad a \leq 0,\quad b \leq 0

In case (1) we also have $a+b \geq 0$ , and the theorem is obvious; in fact,

|a+b| = a + b = |a| + |b|

so that in this case equality holds.

In case (4) we have $a+b \leq 0$ , and again equality holds:

|a+b| = -(a+b) = -a + (-b) = |a| + |b|

In case (2), when $a \geq 0$ and $b \leq 0$ , we must prove that

|a+b| \leq a - b

This case may therefore be divided into two subcases. If $a + b \geq 0$ , then we must prove that

a + b \leq a - b

i.e., b <= -b

which is certainly true since $b < 0$ and hence $-b > 0$ . On the other hand, if $a + b \leq 0$ , we must prove that

-a - b \leq a - b

i.e., -a \leq a

which is certainly true since $a \geq 0$ and hence $-a \leq 0$ .

Finally, note that case (3) may be disposed of with no additional work, by applying case (2) with $a$ and $b$ interchanged.

Reference: Calculus Micheal Spivak. 1. Basic Properties of Numbers