Reverse Triangle Inequality Proof

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.

• proofwiki.org/wiki/Reverse_Triangle_Inequality – dls Apr 2 '12 at 20:15
• 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. – Anonymous Apr 2 '12 at 20:16
• Just replace $d(x,y)$ with $|x-y|$. – dls Apr 2 '12 at 20:21
• 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? – Charlie Parker Feb 14 '18 at 16:24
• 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). – user428487 Mar 27 '18 at 0:50

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

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

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:

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

• Since you have it tagged as real-analyis, this proof assumes real numbers $x,y$. – Aryabhata Apr 2 '12 at 20:20
• OK now I get it - you say: $|y-x|\ge |y|-|x|$ and $|x-y|\ge |x|-|y|$ therefore $|x-y|\ge ||x|-|y||$ – Anonymous Apr 2 '12 at 20:44
• Indeed this shows that the function $x \mapsto |x|$ is Lipschitz continuous with constant 1. – copper.hat Apr 2 '12 at 22:07
• @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. – Aryabhata Feb 15 '18 at 0:45
• @Aryabhata fantastic insight! My sincerely thanks. Wish I could upvote you again. – Charlie Parker Feb 15 '18 at 15:19

Explicitly, we have \begin{align} \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\}\nonumber\\ =&|x-y|.\nonumber \end{align}

WLOG, consider $|x|\ge |y|$. Hence: $$||x|-|y||=||x-y+y|-|y||\le ||x-y|+|y|-|y||=||x-y||=|x-y|.$$

• 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 $\left|a-|y|\right| \leq \left|b-|y|\right|$. – user46234 Sep 12 '18 at 2:38
• Since $|x|\ge |y|$, then $||x|-|y||=|x|-|y|\ge 0$. And we are replacing $|x|$ with the bigger or equal number $|x-y|+|y|$. – farruhota Sep 12 '18 at 2:43

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.

• 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. – RRL Mar 16 '16 at 8:16
• @RRL what is an example where the inequality would fail if the outer absolute value were not present in the reverse triangle inequality? Or is it just a more powerful condition so it is useful to show that? – makansij Jan 23 '19 at 7:39
• @jaja: There isn't any. $|x| = |x-y+y| \leqslant |x-y| + |y| \implies |x| - |y| \leqslant |x-y|$ and switching names $|y| - |x| \leqslant |y -x| = |x- y| \implies |x| - |y| \geqslant -|x-y|$. So we also have $||x|-|y|| \leqslant |x-y|$. My point was that without further work -- like I show here -- you can't immediately conclude that something < |something else| implies |something| < |something else|. – RRL Jan 23 '19 at 7:56
• I see, so then why isn't the reverse triangle inequality stated in its more general form without the outter absolute values? – makansij Jan 24 '19 at 0:47

We can write the proof in a way that reveals how we can think about this problem.

The inequality $$|a|\le M$$ is equivalent to $$-M\le a\le M$$, which is one way to write the following two inequalities together: $$a\le M,\quad a\ge -M\;.$$ Therefore, what we need to prove are (both of) the following: $$|x|-|y|\le |x-y|,\tag{1}$$ $$|x|-|y|\ge -|x-y|\;.\tag{2}$$

On the other hand, the known triangle inequality tells us that "the sum of the absolute values is greater than or equal to the absolute value of the sum": $$|A|+|B|\ge |A+B|\;\tag{3}$$ Observe that there are two (positive) quantities on the left of the $$\ge$$ sign and one of the right. Furthermore, (1) and (2) can be written in such a form easily: $$|y|+|x-y|\ge |x|\tag{1'}$$ $$|x|+|x-y|\ge |y|\tag{2'}$$

Thus, we get (1') easily from (3), by setting $$A=y$$, $$B=x-y$$.

How about (2')? We don't, in general, have $$x+(x-y)=y$$. But wait, (2') is equivalent to $$|x|+|y-x|\ge |y|\tag{2''}$$ because $$|x-y|=|y-x|$$. Now we are done by using (3) again.

$$\left| |x|-|y| \right|^2 - |x-y|^2 = \left( |x| - |y| \right)^2 - (x-y)^2 = |x|^2 - 2|x| \cdot |y| +y^2 - x^2 + 2xy-y^2 = 2 (xy-|xy|) \le 0 \Rightarrow \left| |x|-|y| \right| \le |x-y|.$$

$$"=" \iff xy\ge 0.$$ Q.E.D.

The Triangle Inequality can be proved similarly.

For all $$x,y\in \mathbb{R}$$, the triangle inequality gives $$$$|x|=|x-y+y| \leq |x-y|+|y|,$$$$

$$$$|x|-|y|\leq |x-y| \tag{1}.$$$$ Interchaning $$x\leftrightarrow y$$ gives $$$$|y|-|x| \leq |y-x|$$$$ which when rearranged gives $$$$-\left(|x|-|y|\right)\leq |x-y|. \tag{2}$$$$ Now combining $$(2)$$ with $$(1)$$, gives $$$$-|x-y| \leq |x|-|y| \leq |x-y|.$$$$ This gives the desired result $$$$\left||x|-|y|\right| \leq |x-y|. \blacksquare$$$$