Triangle inequality and its equality

Question

How do I prove this? $$|x+y|=|x|+|y|\Leftrightarrow xy\geq0$$

I tried to use the triangle inequality, but I didn't get so far... Thanks!

Answer 1

The way you can prove this depends on your space. But assuming you are working in $\mathbb R$, the right hand side means that at least one of $x$ or $y$ needs to be zeros. But clearly, the triangle equality holds for $x=y=1$, so that must not be the right space. So that cannot be it.

Is it possible that you are using a vector space, and $x\cdot y$, an inner product? in that case, let $| z|= (z\cdot z)^{1/2}$ and expend the left hand side.

Sorry for the guesswork.

Answer 2

$$|x+y|=|x|+|y| \iff \exists A,B\ge 0 \ \ \ Ax=By$$

This is a consequence of the Cauchy-Schwarz inequality, and is true in every Euclidian (or prehilbertian) space.

Indeed,

$$ |x+y|^2 = |x|^2 + |y|^2 + 2\Re\langle x,y\rangle \le |x|^2 + |y|^2 + 2|\langle x,y\rangle| \leq |x|^2 + |y|^2 + 2|x||y| = (|x|+|y|)^2 $$ now there is equality iff $$ \langle x,y\rangle = |x||y| \iff (x,y) \text{ are on the same line and } \langle x,y\rangle>0 $$ which is equivalent to $$\exists A,B\ge 0 \ \ \ Ax = By $$

Answer 3

The statement simply tells that to add two numbers with the same sign, you have to add their absolute values.