Triangle inequality

The triangle inequality takes a few common forms of which $$|x-z|\le|x-y|+|y-z|$$ is a special case.

Another common way of writing it is $$|a+b|\le |a|+|b|$$ , notice if we set and then we get $$|x-y+y-z|\le|x-y|+|y-z|$$ which is just $$|x-z|\le|x-y|+|y-z|$$

Reverse Triangle Inequality

This is $$|a|-|b|\le|a-b|$$

Proof

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

Note

However we see $$|b|-|a|\le|b-a|$$ but $$|b-a|=|(-1)(a-b)|=|-1||a-b|=|a-b|$$ thus $$|b|-|a|\le|a-b|$$ also.

That is both:

• $$|a|-|b|\le|a-b|$$
• $$|b|-|a|\le|a-b|$$

Full form

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

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

[<collapsible-expand>]Real Analysis

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