## FANDOM

1,168 Pages

The triangle inequality is a statement about the distances between three points: Namely, that the distance from $A$ to $C$ is always less than or equal to the distance from $A$ to $B$ plus the distance from $B$ to $C$ . It can be thought of as "the longest side of a triangle is always shorter than the sum of the two shorter sides".

## Real scalars

For real numbers, the formal statement of the inequality is:

$\forall x,y\in\R:|x+y|\le|x|+|y|$

A corollary of this result, also known as the "reverse triangle inequality", is:

$\forall x,y\in\R:\Big||x|-|y|\Big|\le|x-y|$

### Proof

Proof of the first result is:

As $x\le|x|\ ,\ y\le|y|$ then $x+y\le|x|+|y|$ .

Also $-x\le|x|\ ,\ -y\le|y|$ then $-(x+y)\le|x|+|y|$ .

These two results mean that

$|x+y|=\max\Big\{x+y,-(x+y)\Big\}\le|x|+|y|$

i.e. $|x+y|\le|x|+|y|$ .

Proof of the corollary:

By the first part, $|x|=\Big|(x-y)+ y\Big|\le|x-y|+|y|\iff|x|-|y|\le|x - y|$ .

Also, $|y|=\Big|(y-x)+x\Big|\le|y-x|+|x|\iff|y|-|x|\le|y-x|=|x-y|$

These two results mean that

$\Big||x|-|y|\Big|=\max\Big\{|x|-|y|,-(|x|-|y|)\Big\}\le|x-y|$

i.e. $\Big||x|-|y|\Big|\le|x-y|$

### More than two numbers

The triangle inequality can also be extended to more than two numbers, via a simple inductive proof:

For $n=1$ , clearly $|x_1|\le|x_1|$ . Now suppose that for some $n=k\in\N,\left|\sum_{i=1}^k x_i\right|\le\sum_{i=1}^k|x_i|$ .

Then by the proof above, $\left|\sum_{i=1}^{k+1}x_i\right|=\left|x_{k+1}+\sum_{i=1}^k x_i\right|\le|x_{k+1}|+\left|\sum_{i=1}^k x_i\right|$ .

By the inductive hypothesis we assumed, $|x_{k+1}|+\left|\sum_{i=1}^k x_i\right|\le|x_{k+1}|+\sum_{i=1}^k|x_i|=\sum_{i=1}^{k+1}|x_i|$ .

Therefore by induction, $\forall n\in\N,\left|\sum_{i=1}^n x_i\right|\le\sum_{i=1}^n|x_i|$ .

## Extended Triangle Inequality

The Triangle Inequality could also be used if a triangle is acute, right or obtuse.

### Theorem Statement

If the sum of the squares of any of the 2 sides of the triangle is more than the square of the length of the third side, then the triangle is actue.

If the sum of the squares of two of the sides of the triangle is equal than the square of the length of the third side, then the triangle is right.

I the sum of the squares of two of the sides of the triangle is less than the square of the length of the third side, then the triangle is obtuse.

### Sample Problem

Suppose the shortest side of an obtuse triangle has a length of 10 and the other two sides have lengths of $2a+3$ and $3a+2$ . If $a>0$ , find the range of values for $a$ .

Community content is available under CC-BY-SA unless otherwise noted.