Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
yes you are correct it is similar –  Zoro Dec 17 '13 at 4:50
    
What you have written is true, and your justification is correct! –  angryavian Dec 17 '13 at 4:51
1  
@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

2 Answers 2

up vote 7 down vote accepted

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

share|improve this answer
1  
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 $$\bigl\lvert\abs{x} - \abs{y}\bigr\rvert = \pm(\abs{x} - \abs{y}) \leq \abs{x - y}.$$

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.