Equality holds in triangle inequality iff both numbers are positive, both are negative or one is zero

If $a$ and $b$ are positive, then $|a+b|=a+b=|a|+|b|$. If they are negative, then $|a+b|=-a-b=|a|+|b|$. Suppose one of them is $0$. Without loss of generality suppose $a=0$. Then $|a+b|=|b|=|a|+|b|$.

If none of the three situations occurs, then between $a$ and $b$ one is positive and one negative. Without loss of generality, suppose $a$ is positive. Suppose $|a+b|=|a|+|b|$. If $a+b\geq 0$, then $a+b=a-b$ so that $b=0$, a contradiction. If $a+b<0$, then $-a-b=a-b$ so that $a=0$, a contradiction.

Because $|a+b|$ and $|a|+|b|$ are nonnegative, the inequality $|a+b| \leq |a|+|b|$ is equivalent to $$(|a+b|)^2 \leq (|a|+|b|)^2,$$ which becomes after simplification $$ab \leq |ab|.$$ The equality clearly then holds iff $ab$ is nonnegative.