TITLE: How does the triangle inequality work for $|x-y|$?
QUESTION [12 upvotes]: I know that $|x+y|\leq |x|+|y|$... But is it similar for $|x-y|$? That is, is $|x-y|\leq |x|+|y|$? I ask because of the following:
$x-y=x+(-y)$, so $|x+(-y)|\leq |x|+|-y|=|x|+|y|$
Is it possible that there is a "better" inequality? For example is $|x-y|\leq |x|-|y|$? My textbook only mentions the fact about $|x+y|$ but nothing about any other form.

REPLY [5 votes]: The proof of the "reverse triangle inequality" requested in the other comments: given the basic TE,
$$\def\abs#1{\lvert#1\rvert}\abs{x + y} \leq \abs{x} + \abs{y},$$
subtract $\abs{y}$ from both sides:
$$\abs{x + y} - \abs{y} \leq \abs{x}.$$
Now replace $x + y$ with $x$, and thus $x$ with $x - y$ (if you don't get this: use a new variable $z = x + y$, so $x = z - y$, and then replace $z$ with $x$ later because the names are meaningless):
$$\abs{x} - \abs{y} \leq \abs{x - y}.$$
By symmetry of $x$ and $y$, we also have
$$\abs{y} - \abs{x} \leq \abs{y - x} = \abs{x - y}.$$
Therefore
$$\bigl\lvert\abs{x} - \abs{y}\bigr\rvert = \pm(\abs{x} - \abs{y}) \leq \abs{x - y}.$$