Prove $| |y-x| - |z-y| | \leq |x-z|$

Question

How to prove equation of the form $| |y-x| - |z-y| | \leq |x-z| x,y, z \in \Re$ ?

There are 10(?) cases:
(1) $x>y>z$ : $||y-x|-|z-y|| \le x-z$
(2) $x>z>y$ : $|y-x|-z+y \le x-z$
(3) $y>x>z$ : $|y-x-|z-y|| \le x-z$
(4) $y>z>x$ : $|y-x-|z-y||\le |x-z|$
(5) $z>x>y$ : $||y-x|-z+y|\le|x-z|$
(6) $z>y>x$ : $|y-x-z+y|\le|x-z|$
(7) $x=y>z$ : $|x-x|-|z-x|\le|x-z|$
(8) $x=z>y$ : $|y-x|-|x-y| \le |x-x|$
(9) $y=z>x$ : $|z-x|-|z-z| \le |x-z|$
(10)$x=y=z$ : $||x-x|-|x-x||\le |x-x|$

(1) $||y-x|-|z-y||\le x-z \to ||x-y|-|z-y||\le x-z \to |x-y-z+y|\le x-z \\ \to |x-z|\le x-z \to x-z \le x-z$ proved
(2) $|y-x|-z+y\le x-z \to |y-x|+y\le x \to |x-y|+y\le x \to x \le x$ proved
(3) $|y-x-|z-y|| \le x-z \to |y-x-|y-z|| \le x-z \to |y-x-y+z| \le x-z \to \\|-x+z|\le x-z \to |z-x|\le x-z \to |z-x| \le x-z \to |x-z| \le x-z \to \\ x-z\le x-z $ proved
(4) $|y-x-|z-y|| \le |x-z| \to |y-x-|y-z||\le z-x \to |-x+z| \le z-x \to \\ |-x+z| \le z-x \to |z-x| \le z-x \to z-x \le z-x $ proved (5) $||y-x|-z+y|\le z-x \to |x-y-z+y|\le z-x \to |x-z|\le z-x \to z-x\le z-x$ proved
(6) $|y-x+y-z| \le |x-z| \to |2y-x-z|\le|x-z| \to $ still no idea
(7) $|0-|z-x||\le|x-z| \to |z-x|\le|x-z|$ proved
(8) $|y-x|-|x-y|\le|x-x| \to 0 \le 0$ proved
(9) $|z-x|-|z-z| \le |x-z| \to |z-x|\le|x-z|$ proved
(10)$||x−x|−|x−x||≤|x−x| \to 0\le0$ proved

any ideas how to proceed? what could i do next?

Have you heard of the reverse triangle inequality? $\big\vert|x|-|y|\big\vert\leq|x-y|$. Should reduce the number of cases to check. — csch2, Oct 21 '19 at 12:35

score 4 · Accepted Answer · answered Oct 21 '19 at 12:35

4

By triangle inequality: $$|y-x|=|y-z+z-x|\leq |x-z|+|z-y|$$ and $$|z-y|=|y-x+x-z|\leq |y-x|+|x-z|.$$

answered Oct 21 '19 at 12:35

Floris Claassens

4,611

1

How does that get the desired bound on their difference? – J.G. Oct 21 '19 at 12:42
1

@J.G.: It implies that $-|x-z| \le |y-x| - |z-y| \le |x-z|$, which is exactly the desired estimate. – Martin R Oct 21 '19 at 12:54
Thanks for the help! But I still dont know what to do in Case #6. – VitaCola2010 Oct 21 '19 at 15:27
Case 6 is actually 2 cases, you would have to split it up into $2y<z+x$ and $z+x\leq 2y$. I would suggest sticking with the more general prove I and others provided though. Far less work. – Floris Claassens Oct 21 '19 at 19:18

user · Answer 2 · 2019-10-21T13:59:10.347

2

Recall that by reverse triangle inequality

$$||a|-|b||\le |a-b|$$

indeed

$a,b\ge 0 \implies ||a|-|b||=|a-b|\le |a-b|$
$a,b\le 0 \implies ||a|-|b||=|-a+b|\le |a-b|$
$-a=a'\le 0, b\ge 0 \implies ||a|-|b||=|a'-b|\le |-a'-b|=|a'+b|$
$a\ge 0, b=-b'\le0 \implies ||a|-|b||=|a-b'|\le |a+b'|$

therefore

$$| |x-y| - |z-y| | \leq |(x-y)-(z-y)|=|x-z|$$

edited Oct 21 '19 at 13:59

answered Oct 21 '19 at 12:39

user

154,566

Prove $| |y-x| - |z-y| | \leq |x-z|$

2 Answers2