Proof of absolute value identities $|x+y| = |x| + |y|$ and $|x-y| = |x | - |y|$ How to prove that:
$|x+y| = |x| + |y|$ if and only if $xy \ge0$  
and
$|x-y| = |x | - |y|$ if and only if $xy \le0 $  and $|x| \ge |y|$?
I understand that these have similar approach and that each needs two proofs (that A implies B and that B implies A), but for know I've got as far as just writing $xy \ge 0$ and making different operations with it.
 A: It's 
$$\left(|x+y|\right)^2=\left(|x|+|y|\right)^2$$ or$$x^2+2xy+y^2=x^2+2|xy|+y^2$$ or
$$|xy|=xy,$$ which is $$xy\geq0.$$
Done!
A: It's an "if and only if" sentence, so you prove it by:


*

*Proving that if $xy\geq 0$, then $|x+y| = |x|+|y|$. (you can do this by separating the cases where $x>0$ and $x<0$ and $x=0$)

*Proving that if $|x+y|=|x|+|y|$, then $xy\geq 0$. You can most easily prove this using the reverse, i.e. proving that if $xy<0$, then $|x+y|\neq|x|+|y|$, and again, you can prove this by separating the cases.


Same for the other one.
A: As for the first:
If $xy<0$ it means that only one if them is negative, so $|y+x|$ is or $|y+(-x)|$ or $|(-y)+x|$ which is not equal to $|y|+|x|$, now if $xy\ge 0$ than both of them are positive or both negative, if both negative you have $|(-y)+(-x)|=|-(y+x)|=|y|+|x|$ and if both positive than $|(y)+(x)|=|y+x|=|y|+|x|$. Now if you want it to be "if and only if" you need to show that also if $|y+x|=|y|+|x|$ then $xy\ge 0$
You can do similar thing to the second one
