Prove that triangle inequality $|a + b| \le |a| + |b|$ holds when $(a + b) \ge 0, a \ge 0, b < 0$ This is what I did: 
$a + b \ge  0 \rightarrow |a + b| = a + b$
$a \ge 0 \rightarrow |a| = a$ 
$a + b \le a + |b|$
$b \le |b|$
Which is true $\forall b$. Is this a formal enough way of proving the original statement under given conditions?
 A: This looks formal enough, yes. Just make sure to mention that the line
$$a+b\leq a+|b|$$
and the line
$$b\leq |b|$$
are equivalent (not only that one follows from the other). You can then even use the fact that $b<0$ and $0<|b|$ to prove that last line.
A: It is easier:
We have $|x|≥x$ and $|y|≥y$ and therefore: $|x|+|y|≥x+y$
Similarly: $|x|=|-x|≥-x$ and $|y|=|-y|≥-y$ and thus $|x|+|y|≥-x-y=-(x+y)$
Combining yields:
$$
|x|+|y|≥\max{(x+y,-x-y)}=|x+y|
$$
A: The simplest proof, in my opinion, is this: 
Observing that both members of the inequality are non-negative, we can square both sides. Thus, using $\lvert x\rvert ^2=x^2$ and $\lvert a\rvert\cdot\lvert b\rvert= \lvert ab\rvert$:
\begin{align*}
\lvert a+b\rvert\le \lvert a\rvert+\lvert b\rvert&\iff (a+b)^2\le a^2+2\lvert ab\rvert+b^2\iff ab\le\lvert ab\rvert,
\end{align*}
which is true whatever the sign of $ab$.
A: The easiest proof of the triangle inequality for real numbers, in my opinion, does not involve the sign of the numbers at all. Since $a\leqslant |a|$ and $b\leqslant|b|$, we have
$$a+b\leqslant |a| + |b|.\tag1 $$
Since $-a\leqslant|a|$ and $-b\leqslant|b|$, we have
$$(-a)+(-b) = -(a+b) \leqslant |a| + |b|.\tag2$$
From $(1)$ and $(2)$ it follows by definition that
$$|a+b| \leqslant |a| + |b|. $$
