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I've seen the full proof of the Triangle Inequality \begin{equation*} |x+y|\le|x|+|y|. \end{equation*} However, I haven't seen the proof of the reverse triangle inequality: \begin{equation*} ||x|-|y||\le|x-y|. \end{equation*} Would you please prove this using only the Triangle Inequality above?

Thank you very much.

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    http://www.proofwiki.org/wiki/Reverse_Triangle_Inequality2012-04-02
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    I've seen this proof, however it's too advanced for me as it involves metric spaces - I'd like a simple proof using the known and simple triangle inequality I wrote in the question, thanks.2012-04-02
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    Just replace $d(x,y)$ with $|x-y|$.2012-04-02
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    this inequality has always bothered me, its never really been an intuitive thing that I would come up with and every proof just seems like symbol crunching. What is the main concepts going on in this proof?2018-02-14
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    If you think about $x$ and $y$ as points in $\mathbb{C}$, on the left side you're keeping the distance of both the vectors from 0, but making them both lie on the positive real axis (by taking the norm) before finding the distance, which will of course be less than if you just find the distance between them as they are (when they might be opposite each other in the complex plane).2018-03-27

4 Answers 4

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$$|x| + |y -x| \ge |x + y -x| = |y|$$

$$|y| + |x -y| \ge |y + x -y| = |x|$$

Move $|x|$ to the right hand side in the first inequality and $|y|$ to the right hand side in the second inequality. We get:

$$|y -x| \ge |y| - |x|$$

$$|x -y| \ge |x| -|y|.$$

From absolute value properties we know that $|y-x| = |x-y|$ and if $t \ge a$ and $t \ge −a$ then $t \ge |a|$.

Combining these two facts together we get the reverse triangle inequality:

$$|x-y| \ge \bigl||x|-|y|\bigr|.$$

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    Since you have it tagged as real-analyis, this proof assumes real numbers $x,y$.2012-04-02
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    Yes, I'm assuming real numbers x,y. Could you please finish the proof and show the reverse inequality? Thanks a lot!2012-04-02
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    @Anonymous: I suggest you try finishing it yourself first.2012-04-02
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    I tried finishing it myself, however all I could come up with is substracting both of the inequalities which didn't get me to the final inequality :\2012-04-02
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    Here is another hint: If $x \ge a$ and $x \ge -a$, then can you say $x \ge |a|$? Also, is it true that $|x| = |-x|$?2012-04-02
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    yes to both of the questions.2012-04-02
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    @Anonymous: Does that help you finish it?2012-04-02
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    Unfortunately, no :(2012-04-02
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    Another hint: Try moving the $|x|$ in the first inequality and the $|y|$ in the second inequality to the right side...2012-04-02
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    OK now I get it - you say: $|y-x|\ge |y|-|x|$ and $|x-y|\ge |x|-|y|$ therefore $|x-y|\ge ||x|-|y||$2012-04-02
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    @Anonymous: Exactly!2012-04-02
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    @Anonymous: You are welcome.2012-04-02
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    Indeed this shows that the function $x \mapsto |x|$ is Lipschitz continuous with constant 1.2012-04-02
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    @user27978: Right!2012-04-02
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    this inequality has always bothered me, its never really been an intuitive thing that I would come up with and every proof just seems like symbol crunching. What is the main concepts going on in this proof?2018-02-14
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    @CharlieParker: Intuitively, if $x$ and $y$ have the same sign then $||x| - |y||$ is the same as $|x - y|$ (the distance between $x$ and $y$ when plotted on the real line). If they are different, then the distance between $x$ and $y$ is larger than the distance between $x$ (say $x \ge 0$) and the reflected version of $y$, i.e. $|y|$. Might help if you draw it out.2018-02-15
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    @Aryabhata fantastic insight! My sincerely thanks. Wish I could upvote you again.2018-02-15
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    @CharlieParker: Glad it helped!2018-02-15
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Given that we are discussing the reals, $\mathbb{R}$, then the axioms of a field apply. Namely for :$x,y,z\in\mathbb{R}, \quad x+(-x)=0$; $x+(y+z)=(x+y)+z$; and $x+y=y+x$.

Start with $x=x+0=x+(-y+y)=(x-y)+y$.

Then apply $|x| = |(x-y)+y|\leq |x-y|+|y|$. By so-called "first triangle inequality."

Rewriting $|x|-|y| \leq |x-y|$ and $||x|-y|| \leq |x-y|$.

The item of Analysis that I find the most conceptually daunting at times is the notion of order $(\leq,\geq,<,>)$, and how certain sentences can be augmented into simpler forms.

Hope this helps and please give me feedback, so I can improve my skills.

Cheers.

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    You can't immediately conclude that $||x|-|y|| \leqslant |x-y|$ from $|x|-|y| \leqslant |x-y|$ since you don't know signs at this point. Note that $-2 < 1$ does not imply $|-2| < 1$. But you can fix by switching variable labels and showing $|y| - |x| \leqslant |x-y|$ which implies $|x| -|y| \geqslant -|x-y|$. Fill in the details and I'll upvote.2016-03-16
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WLOG, consider $|x|\ge |y|$. Hence: $$||x|-|y||=||x-y+y|-|y||\le ||x-y|+|y|-|y||=||x-y||=|x-y|.$$

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    Is it obvious that the inequality in the middle holds? For example, I don't think it's generally true that if $a\leq b$, then $\left|a-|y|\right| \leq \left|b-|y|\right|$.2018-09-12
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    Since $|x|\ge |y|$, then $||x|-|y||=|x|-|y|\ge 0$. And we are replacing $|x|$ with the bigger or equal number $|x-y|+|y|$.2018-09-12
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Explicitly, we have \begin{equation} \bigl||x|-|y|\bigr| = \left\{ \begin{array}{ll} |x-y|=x-y,&x\geq{}y\geq0\\ |x-y|=-x+y=-(x-y),&y\geq{}x\geq0\\ |-x-y|=x+y\leq-x+y=-(x-y),&y\geq-x\geq0\\ |-x-y|=-x-y\leq-x+y=-(x-y),&-x\geq{}y\geq0\\ |-x+y|=-x+y=-(x-y),&-x\geq-y\geq0\\ |-x+y|=x-y,&-y\geq-x\geq0\\ |-x+y|=-x-y\leq{}x-y,&-y\geq{}x\geq0\\ |-x+y|=x+y\leq{}x-y,&x\geq-y\geq0 \end{array} \right\} =|x-y|.\nonumber \end{equation}