How would I prove $|x + y| \le |x| + |y|$? How would I write a detailed structured proof for:

for all real numbers $x$ and $y$, $|x + y| \le |x| + |y|$

I'm planning on breaking it up into four cases, where both $x,y < 0$, $x \ge 0$ and $y<0$, $x<0$ and $y \ge0$, and $x,y \ge 0$. But I'm not sure how I'd go about writing it formally.
Thanks!
 A: You are absolutely on the right track. I'll model one case for you, and you can try the other cases on your own.
Case 1: $x,y\geq 0$. Then $x+y\geq 0$, so $|x+y|=x+y$. Similarly, $|x|=x$ because $x\geq 0$, and $|y|=y$ because $y\geq 0$. Thus $|x+y|=x+y=|x|+|y|$.
A: Here is an alternative proof, without any case distinctions, using the definition
$$
|p| = p \max -p
$$
so that we can use the properties of $\;\max\;$, which are simpler than those of $\;|\phantom p|\;$.  In this case we start at the right hand side of the equation, which seems to be the most complex side, and calculate for every $\;x,y\;$:
\begin{align}
& |x| + |y| \\
= & \;\;\;\;\;\text{"the above definition, twice"} \\
& (x \max -x) + (y \max -y) \\
= & \;\;\;\;\;\text{"$\;+\;$ distributes over $\;\max\;$"} \\
& (x + (y \max -y)) \max (-x + (y \max -y)) \\
= & \;\;\;\;\;\text{"$\;+\;$ distributes over $\;\max\;$, twice more; $\;\max\;$ is associative"} \\
& (x+y) \max (x-y) \max (-x+y) \max (-x-y) \\
\geq & \;\;\;\;\;\text{"$\;p \max q \geq p\;$, twice"} \\
& (x+y) \max (-x-y) \\
= & \;\;\;\;\;\text{"the above definition"} \\
& |x + y| \\
\end{align}
A: You might consider the following:
Consider squaring both sides since both sides are non negative (you don't have to worry about sign flips). Canceling yields
$$xy \leq |x||y|$$
which is true.
