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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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    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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    @CharlieParker: Glad it helped!2018-02-15
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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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    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
6

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}

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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.