Prove that for all real numbers x and y we have $| |x| - |y| | \leq |x-y|$.

Question

This is my second attempt at proving this inequality, the first one went terribly so hopefully I did a better job this time.

Firstly we are given as fact (by the examiner). $||x|-|y|| = |x| - |y| $ for all $|x| \geq |y|$ and $||x|-|y|| = |y| - |x|$ for all $|x| \leq |y|$.

Let...

$$ |x| \leq |y| \iff |x|-|y| \leq 0 \leq|x-y| $$

Let..

$$ |y| \leq |x| \iff 0\leq|x| - |y| = |x-y| = |y-x| \leq|x-(-y)| = |(-y)-x|| $$

Therefore for any real number $x$ and $y$ the inequality $||x|-|y|| \leq |x-y|$ holds.

Any feedback would be much appreciated.

Not valid. When $|x| \leq |y|$ you have to prove $|y|-|x| \leq |x-y|$ also. — Kavi Rama Murthy, Apr 16 '20 at 12:34

score 1 · Accepted Answer · answered Apr 16 '20 at 13:15

1

Your proof is incorrect. In the first case, you actually have to show $|y|-|x|\leq |x-y|.$ In the second case, you write $|x|-|y|=|x-y|,$ which is incorrect.

A correct proof can be found here, or here or for an arbitrary metric.

answered Apr 16 '20 at 13:15

Sahiba Arora

10,847

Prove that for all real numbers x and y we have $| |x| - |y| | \leq |x-y|$.

1 Answers1