Proof of absolute value identities $|x+y| = |x| + |y|$ and $|x-y| = |x | - |y|$

Question

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.

Answer 1

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!

Answer 2

It's an "if and only if" sentence, so you prove it by:

1. 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 $x=0$)
2. 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.

Answer 3

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