Here are my "proofs" for Spivak's Calculus Chapter 1 Problem 12. I am new to this level of rigour and I am attempting to intimate myself with more advanced topics of mathematics to prepare for next year. I apologize in advance, as these so-called "proofs" are not likely to be nearly as rigorous as they should be. Any assistance on how to write the proofs better or any critiques on faulty logic would be greatly appreciated.
12) v) Prove that $|x|-|y|\leq|x-y|$ (A very short proof is possible if you write it the right way.)
(This is where I start feeling especially uncomfortable with my answers.) My proof is:
By (iv), $|x-y|+|y|\geq|(x-y)+y|$
$\therefore$
$|x-y|+|y|\geq|x|$, and $|x|-|y|\leq|x-y|$, which is what we wished to prove.
12) vi) Prove that $|(|x|-|y|)|\leq|x-y|$ (Why does this follow immediately from (v)?)
My proof is:
$|(|x|-|y|)|\leq|x-y|$ has two primary cases, $|x|-|y|\leq|x-y|$ and $-(|x|-|y|)=|y|-|x|\leq|x-y|$
The first case follows immediately from (v), and the second case follows when $x$ and $y$ are switched in (v). ($|x-y|=|y-x|$).
12) vi) Prove $|x+y+z|\leq|x|+|y|+|z|$ and indicate when equality holds.
My proof is:
$|x+y+z|=|x-(-y-z)|$, which by (iv)$\leq|x|+|-y-z|=|x|+|y+z|$
$\therefore$
$|x|+|y+z|=|x|+|y-(-z)|\leq|x|+|y|+|-z|=|x|+|y|+|z|$, which is what we wished to prove.
By (iv), equality holds when $|x|+|y|=|x+y|$, which is when $x$ and $y$ have the same sign. By extension $|x+y+z|=|x|+|y|+|z|$ when $z$ has the same sign as $x$ and $y$. (With the exception of the cases where one variable equals zero.)
These all look great! It doesn't look like you need to stick in the minus signs on $y$ and $z$ at the beginning of vi), though.