How do we show that equality holds in the triangle inequality $|a+b|=|a|+|b|$ iff both numbers are positive, both are negative or one is zero? I already showed that equality holds when one of the three conditions happens.
$\endgroup$ 24 Answers
$\begingroup$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.
$\endgroup$ 4 $\begingroup$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.
$\endgroup$ $\begingroup$If we have $$|a + b| = |a| + |b|$$
Then we have two cases. First $$a + b = |a| + |b| \implies a-|a| =|b|-b$$ Both sides in the above are either simultaneously zero (in which $a = |a|$ and $b = |b|$) or simultaneously not zero, in which ($a \neq |a|$ and $b \neq |b|$). The first case is simultaneously positive and the second implies $|a| = |b| = 0$.
Similarly for the other case $$ -a - b = |a| + |b| \implies -|a|-a = b+|b|$$ in which the same analysis applies.
$\endgroup$ $\begingroup$If we have $|a+b | = | a | + | b|$, squaring both sides we obtain$$|a+b|^2 = (|a| + |b|)^2$$
$$\iff a^2+2ab+b^2 = |a|^2 + 2|a||b| +|b|^2 = a^2 + 2|a||b| + b^2.$$
Finally, $$ab = |a||b| \ge 0 .$$
So that, $|a+b| = |a| + |b|$ iff $ab \ge 0$.
$\endgroup$
