Difference between revisions of "Triangle inequality"

Revision as of 12:06, 14 April 2015

The triangle inequality takes a few common forms, for example: [math]d(x,z)\le d(x,y)+d(y,z)[/math] (see metric space) of which [math]|x-z|\le|x-y|+|y-z|[/math] is a special case.

Another common way of writing it is [math]|a+b|\le |a|+|b|[/math], notice if we set [ilmath]a=x-y[/ilmath] and [ilmath]b=y-z[/ilmath] then we get [math]|x-y+y-z|\le|x-y|+|y-z|[/math] which is just [math]|x-z|\le|x-y|+|y-z|[/math]

Definition

The triangle inequality is as follows:

• [math]|a+b|\le |a|+|b|[/math]

Proof

Style: case analysis

TODO: Take time to write it out

Reverse Triangle Inequality

This is [math]|a|-|b|\le|a-b|[/math]

Proof

Take [math]|a|=|(a-b)+b|[/math] then by the triangle inequality above:
[math]|(a-b)+b|\le|a-b|+|b|[/math] then [math]|a|\le|a-b|+|b|[/math] clearly [math]|a|-|b|\le|a-b|[/math] as promised

Note

However we see [math]|b|-|a|\le|b-a|[/math] but [math]|b-a|=|(-1)(a-b)|=|-1||a-b|=|a-b|[/math] thus [math]|b|-|a|\le|a-b|[/math] also.

That is both:

• [math]|a|-|b|\le|a-b|[/math]
• [math]|b|-|a|\le|a-b|[/math]

Full form

There is a "full form" of the reverse triangle inequality, it combines the above and looks like: [math]|a-b|\ge|\ |a|-|b|\ |[/math]

It follows from the properties of absolute value, I don't like this form, I prefer just "swapping" the order of things in the abs value and applying the same result