real analysis – Reverse Triangle Inequality Proof

The Question :

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I’ve seen the full proof of the Triangle Inequality
However, I haven’t seen the proof of the reverse triangle inequality:
Would you please prove this using only the Triangle Inequality above?

Thank you very much.

The Question Comments :
  • 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.
  • Just replace $d(x,y)$ with $|x-y|$.
  • 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?
  • 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).

The Answer 1

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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:

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

The Answer 2

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

The Answer 3

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Explicitly, we have

The Answer 4

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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.


The Answer 5

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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.

The Answer 6

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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

The Answer 7

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

$”=” \iff xy\ge 0.$ Q.E.D.

The Triangle Inequality can be proved similarly.

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