How does the triangle inequality work for $|x-y|$?

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.

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yes you are correct it is similar – AbKDs Dec 17 '13 at 4:50
What you have written is true, and your justification is correct! – angryavian Dec 17 '13 at 4:51
@agent154 : no one seems to have pointed out that your inequality $|x-y| \leq |x|-|y|$ is in general false: take $x=0$ and $y=1$. – Stefan Smith Dec 17 '13 at 6:10

Indeed, you are correct in your intuition. Moreover, we also have $$|x-y|\ge \left|\; |x|-|y|\; \right|$$

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This is the version that has always been more helpful for me; you get to preserve the sign which is generally a thing you want (or at least I have :P). – Eric Stucky Dec 17 '13 at 5:08
I have here that $||x|-|y||\leq|x+y|\leq |x|+|y|$... I was able to find several proofs of the "reverse triangle inequality", but they all start off with $|x-y|$ instead of $|x+y|$ like in the upper and lower bounds given. How does this proof translate from $|x|-|y|\leq |x-y|$ to $||x|-|y||\leq |x+y|$? – agent154 Dec 17 '13 at 5:28

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 $$\bigl\lvert\abs{x} - \abs{y}\bigr\rvert = \pm(\abs{x} - \abs{y}) \leq \abs{x - y}.$$

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But my question, though, is how $||x|-|y||\leq|x+y|$? My textbook says (without proving) that $||x|-|y||\leq|x+y|\leq |x|+|y|$. – agent154 Dec 17 '13 at 5:52
Nevermind - I found a proof that shows the inequality using $|x+y|$ – agent154 Dec 17 '13 at 6:02
You are aware that replacing $y$ by $-y$ everywhere gives this to you? – Ryan Reich Dec 18 '13 at 2:09
Yes, that's what I ended up finding. Never occured to me – agent154 Dec 18 '13 at 2:12