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

Question

$||x|-|y|| \leq |x-y|$ when $(x,y \in R^k)$

In Principles of MA(Rudin), the author said one sees easily that

$||x|-|y|| \leq |x-y|$ when $(x,y \in R^k)$ (p.88, Rudin)

from the triangle inequality. But I'm not sure how to use the triangle inequality to show this. Can you help me show this?

Answer 1

$|x|=|(x-y)+y|\le|x-y|+|y|$ so $|x|-|y|\le|x-y|$.

$|y|=|(y-x)+x|\le|x-y|+|x|$ so $|y|-|x|\le|x-y|$.

Altogether,

$||x|-|y||\le|x-y|$

Answer 2

Using the triangle inequality, you obtain:

$$|x - y| \geq |x| - |y|$$

and

$$|y - x| \geq |y| - |x|.$$

Since $|x-y| = |y-x|$, you have that

$$-|x-y| \leq |x| - |y| \leq |x-y|$$

which gives the result.

Answer 3

$$|a|+|b|\ge |a+b|$$ Take $a=x-y, b=y$. Then take $a=y-x, b=x$.

Answer 4

Hint: You can write $|x| = |x-0|$