The triangle inequality

Squaring. Both sides are nonnegative, so it suffices to compare their squares:
$$|a+b|^2=(a+b)^2=a^2+2ab+b^2,$$
$$(|a|+|b|)^2=a^2+2|ab|+b^2.$$
Since $ab\le|ab|$, we get $(a+b)^2\le(|a|+|b|)^2$, hence $|a+b|\le|a|+|b|$.

Equality case. Equality holds iff $ab=|ab|$, i.e. $ab\ge0$: the two numbers have the same sign (or one is zero).

Remark. The triangle inequality generalizes to any number of terms: $|a_1+\cdots+a_n|\le|a_1|+\cdots+|a_n|$.