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I am really stuck in the following:

I want to show that for all $\alpha,\beta,\gamma\in\mathbb{R}$ the following is true:

$$|\alpha+\beta-\gamma|+|\alpha+\gamma-\beta|+|\beta+\gamma-\alpha|\ge|\alpha|+|\beta|+|\gamma|$$ I know that I have to proof by cases.

So if I consider $\alpha,\beta,\gamma>0$ don't I have to consider in this case $\alpha+\beta\ge\gamma$ and $\alpha+\beta\le\gamma$, too?

Anybody could help with this inequality? Thanks a lot!

1 Answers 1

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Multiply by 2 and use triangle inequality.

As in: $|\alpha +\beta - \gamma|+|\alpha + \gamma - \beta| \geq |\alpha +\beta - \gamma +\alpha + \gamma - \beta|=2|\alpha|$; similarly for all cyclic permutations and then summ up.

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    Thaks for your comment.2012-11-13
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    @Max well I am just wondering when is it an equality? It's obvious if $\alpha=\beta=\gamma$ but what about other cases? (e.g. 1,2,3)2012-11-22
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    @sheldoor You just need all triangle inequalities to be equalities, that is all $a+b-c$ of the same sign (for $a,b,c$ a permuatation of $\alpha, \beta, \gamma$). This will happen if they are all positive and sides of a triangle, or if they are all negative and $-\alpha, -\beta, -\gamma$ are sides of a triangle.2012-11-22