Proof verification $|x+y| \leq |x| + |y|$ Here's my proof:
Suppose $|x+y|\leq |x|+|y|$
we know that   $-(|x|+|y|)\leq|x+y|\leq|x|+|y|$
1st case: When $x$ and $y$ are negative we get $-(x+y)\leq -(x+y)\leq x+y$
. If we divide by $x+y$ we get $-1\leq-1 \leq1 $ which is true.
2nd case: When $x$ and $y$ are positive by the same reasoning we get $-1\leq 1\leq 1$ which is true.
so from the two cases we conclude that $|x+y|\leq|x|+|y|$
Am I missing something ?
 A: *

*If $x,y\ge 0$, then $|x+y| = x+y = |x|+|y|$.

*If $x,y\le 0$, then $x+y \le 0$ and thus $|x+y| = -(x+y) = (-x)+(-y) = |x|+|y|$.

*Let $x\le 0\le y$. Then $x+y = |y|-|x|$.
3.1. If $|y|-|x|\ge 0$, then $|x+y| = |y|-|x|\le |y|+|x|$.
3.2. If $|y|-|x| < 0$, then $|x+y| = -(|y|-|x|) = |x|-|y|\le |x|+|y|$.

*If $y\le 0\le x$, then this is 3. with $x$ and $y$ swapped. So, we get the same result.
A: Yes, you are missing some cases and you have made some mistakes so far.
Note that if  both x and y are negative then $$-(|x|+|y|)\leq|x+y|\leq|x|+|y|$$  is equivalent to  
$$ x+y\le -x-y\le -x-y $$ and if they are both positive it is equivalent to $$-x-y\le x+y\le x+y $$
And you have not considered  the case where one is positive and the other one is negative. 
A: Let $a= \min(|x|,|y|)$ and $b=\max(|x|, |y|)$.
Then  $0\le a \le b$ and $0 \le b-a \le a+b$
If $x$ and $y$ are both non-negative or both negative then: $|x+y|=|\pm(a+b)|=a+b =|x|+|y|$.
$x$ and $y$ are "mixed" that is one is negative and one is non-negative then: $|x+y| = |\pm a \mp b|=b-a \le a+b = |x| + |y|$.
And that's it.
