The Triangle Inequality

Theorem

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

Proof

We will consider 4 cases:

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

so that in this case equality holds.

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

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

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

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

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