3
$\begingroup$

Assume $a, b \in \mathbb{R}$. Show that $$|a+b|=|a|+|b|\Leftrightarrow ab\geq0$$

The triangle inequality says that for $a,b\in\mathbb{R}$, $|a+b|\leq |a|+|b|$.

So I belive I can say $$|a+b|-|a|-|b|\leq 0$$

I am not quite sure where to go from there, though, or if I am on the right track.

$\endgroup$
  • $\begingroup$ What's a more wordy way of saying $ab \geq 0$ ? $\endgroup$ – pjs36 Sep 18 '15 at 2:27
9
$\begingroup$

By squaring the two sides,

$$ |a+b|=|a|+|b|\Leftrightarrow a^2 + 2ab + b^2 = a^2 + 2|ab| + b^2.$$

Thus $$ |a+b|=|a|+|b|\Leftrightarrow ab = |ab| \Leftrightarrow ab\geq 0.$$

$\endgroup$
  • $\begingroup$ I love when these things don't become case-seeking. +1 $\endgroup$ – Aloizio Macedo Sep 18 '15 at 3:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.