Proof of triangle inequality I understand intuitively that this is true, but I'm embarrassed to say I'm having a hard time constructing a rigorous proof that $|a+b| \leq |a|+|b|$. Any help would be appreciated :)

A simple proof of the triangle inequality that is complete and easy to understand (there are more cases than strictly necessary; however, my goal is clarity, not conciseness).
Prove the triangle inequality $| x | + | y| ≥ | x + y|$.
Without loss of generality, we need only consider the following cases:


*

*$x = 0$

*$x > 0, y > 0$

*$x > 0, y < 0$


Case $1$. Suppose $x = 0$. Then we have
$| x| = 0$
$| x| + | y| = 0 + | y| = | y|$
Thus $| x| + | y| = | x + y|$.
Case $2$. Suppose $x > 0, y > 0$. Then, since $x + y > 0$, we have
$| x| = x$
$| y| = y$
$| x| + | y| = x + y$
$| x + y| = x + y$
Thus $| x| + | y| = | x + y|$.
Case $3$. Suppose $x < 0, y < 0$. Then, since $x + y < 0$, we have
$| x| = −x$
$| y| = −y$
$| x| + | y| = (−x) + (−y)$
$| x + y| = −(x + y) = (−x) + (−y)$
Thus $| x| + | y| = | x + y|$.
Case $4$. Suppose $x > 0, y < 0$. Then we have
$| x| = x$
$| y| = −y$
$| x| + | y| = x + (−y)$
We must now consider three cases:
a. $x + y = 0$
b. $x + y > 0$
c. $x + y < 0$
Case $4a$. Suppose $x + y = 0$. Then we have
$| x + y | = |0| = 0$
Since $y < 0$, it follows that $−y > 0$ and thus $x + (−y) > 0 + (-y) = -y > 0$. 
Therefore, since $| x| + | y| = x + (−y)$, we must have $| x| + | y| > | x + y|$.
Case $4b$. Suppose $x + y > 0$. Then we have
$| x + y| = x + y$
Since $y < 0$, it follows that $−y > 0 > y$ and thus $x + (−y) > x + y$.
Therefore, since $| x| + | y| = x + (−y)$, we must have $| x| + | y| > | x + y|$.
Case $4c$. Suppose $x + y < 0$. Then we have
$| x + y| = −(x + y) = (−x) + (−y)$
Since $x > 0$, it follows that $x > 0 > −x$ and thus $x + (−y) > (−x) + (−y)$.
Therefore, since $| x| + | y| = x + (−y)$, we must have $| x| + | y| > | x + y|$.
This concludes the proof.
