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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!

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  • $\begingroup$ ?​​​​​​​​​​​​​​​ $\endgroup$ – user2345215 Mar 23 '14 at 23:12
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    $\begingroup$ What is x,y and $|.|$? $\endgroup$ – BoZenKhaa Mar 23 '14 at 23:13
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    $\begingroup$ $|1+2|=|1|+|2|$ and $2\times 1\neq 0$ $\endgroup$ – npisinp Mar 23 '14 at 23:15
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    $\begingroup$ It isn't true.. $\endgroup$ – Seth Mar 23 '14 at 23:15
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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.

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

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

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 $$\exists A,B\ge 0 \ \ \ Ax = By $$

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