1. ## Absolute value proof

Can someone help me to understand this proof?

I have to prove that:
$\displaystyle \left| {\left| x \right| - \left| y \right|} \right| \le \left| {x - y} \right|$

Proof from the book goes like this:
1) let be $\displaystyle x - y = z \Rightarrow x = y + z$
2) then is $\displaystyle \left| x \right| = \left| {y + z} \right| \le \left| y \right| + \left| z \right|$
3) then $\displaystyle \left| x \right| \le \left| y \right| + \left| z \right| \Rightarrow \left| z \right| \ge \left| x \right| - \left| y \right| \Rightarrow \left| {x - y} \right| \ge \left| x \right| - \left| y \right|$
4) same way we get $\displaystyle \left| {y - x} \right| \ge \left| y \right| - \left| x \right|$
5) since also $\displaystyle \left| {x - y} \right| = \left| {y - x} \right|{\rm{ and }}\left| {\left| x \right| - \left| y \right|} \right| = \left| {\left| y \right| - \left| x \right|} \right|$
6) we prove that $\displaystyle \left| {x - y} \right| \ge \left| {\left| x \right| - \left| y \right|} \right|$

I understand everything except step 6).
Can someone explain me that?

2. Originally Posted by OReilly
6) we prove that $\displaystyle \left| {x - y} \right| \ge \left| {\left| x \right| - \left| y \right|} \right|$
(I assume these are real numbers)

There are two possibilities for here, on the right side.
Either, $\displaystyle |x|-|y|\geq 0$ or $\displaystyle |x|-|y|<0$
If the first case, then by the definition of absolute value,
$\displaystyle ||x|-|y||=|x|-|y|$ then by step 3 we know this is true.
If the second case, thenby the definition of aboluste vale,
$\displaystyle ||x|-|y||=|y|-|x|$ then by step 4 we know it is true too.

3. Thanks!