The triangle inequality

1. Introduction

The single most important inequality in analysis is the triangle inequality, and it will be used a lot throughout this course. Later on it becomes the main building block for a more general theory of analysis that you learn about when you study metric spaces.

The triangle inequality concerns distance between points and says that the straight line distance between A and B is less than the sum of the distances from A to C and from C to B . It is very much part of our everyday intuition about distances and easy to remember. It is, however, very useful.

2. Distances in the reals

Given real numbers x and y the value | x - y | represents the distance along the numberline from x to y . By definition, it is equal to x - y if x y and to y - x if y > x . So | x - y | = | y - x | . We will sometimes denote this distance by d ( x , y ) or d ( y , x ) .

The triangle inequality.

For all x , y , z we have

d ( x , y ) d ( x , z ) + d ( z , y )

Proof.

This is just by looking at all the cases.

Subproof.

Case 1: x y . Then d ( x , y ) = | x - y | = x - y . There are three subcases. They are all very similar.

Subproof.

Case 1a: z x , so d ( x , z ) = | x - z | = z - x 0 and d ( z , y ) = | z - y | = z - y x - y = d ( x , y ) giving d ( x , z ) + d ( z , y ) d ( x , y ) .

Subproof.

Case 1b: x > z y , so d ( x , z ) = | x - z | = x - z and d ( z , y ) = | z - y | = z - y giving d ( x , z ) + d ( z , y ) = x - z + z - y = x - y = d ( x , y ) .

Subproof.

Case 1c: y > z , so d ( x , z ) = | x - z | = x - z x - y and d ( z , y ) = | z - y | = y - z 0 giving d ( x , z ) + d ( z , y ) = x - y = d ( x , y ) .

Also:

Subproof.

Case 2: x < y . Then d ( x , y ) = | x - y | = y - x . There are three subcases.

Subproof.

Case 2a: z y , so d ( x , z ) = | x - z | = z - x d ( x , y ) and d ( z , y ) = | z - y | = z - y 0 giving d ( x , z ) + d ( z , y ) d ( x , y ) .

Subproof.

Case 2b: y > z x , so d ( x , z ) = | x - z | = z - x and d ( z , y ) = | z - y | = y - z giving d ( x , z ) + d ( z , y ) = z - x + y - z = y - x = d ( x , y ) .

Subproof.

Case 2c: x > z , so d ( x , z ) = | x - z | = x - z 0 and d ( z , y ) = | z - y | = y - z y - x giving d ( x , z ) + d ( z , y ) = y - x = d ( x , y ) .

The triangle inequality in takes the form

| x - y | | x - z | + | z - y |

Note the , the + sign and the introduced intermediate point z on the right. By rearranging we have | x - y | - | z - y | | x - z | , or

| x - z | | x - y | - | y - z |

which also has an intermediate point y added on the right in the same way, but the has changed to and the + has changed to - . Both forms are equally useful. I find the first easy to remember; the hints here will help you remember the second just as easily.

We can also write the triangle inequality in in the form

| x + y | | x | + | y |

To derive this from the other versions just note that

| x - ( - y ) | | x - 0 | + | 0 - ( - y ) |

Hence the result. Its alternative form,

| x + y | | x | - | y |

can be derived in a similar way. If we switch - y for y we can even get

| x - y | | x | + | y |

and

| x + y | | x | - | y |

which can also be useful, especially if we don't know if y is negative or positive.