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I have just proved $|x+y-z|+|x-y+z|+|-x+y+z| \geq |x| + |y| + |z|$ for all $x,y,z\in\mathbb R$ and I want now to find out when this is an equality. I just know the case $x=z=y$ but I've tried $1,2$ and $3$ for $x,y,z$ and it fits. So how do I get the other cases? And how do you prove it?

Proof:

$2|x+y-z|+2|x-y+z|+2|-x+y+z|$

$=|x+y-z|+|x-y+z|+|x-y+z|+|-x+y+z|+|-x+y+z|+|x+y-z|$

$\geq |x+y-z+x-y+z|+|x-y+z-x+y+z|+|-x+y+z+x+y-z|$

$= 2|x| + 2|y| + 2|z|$

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    @HagenvonEitzen proof added. so my guess would be that $x+y-z,x-y+z.-x+y+y$ don't have different signs but how do you prove it?2012-11-22

1 Answers 1

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Looking at your proof, we see that the second $\ge$ is in fact always just $=$.

The second inequality holds because $|x+y-z|+|x-y+z|\ge |(x+y-z)+(x-y+z)|$ (and symmetric variants) by triangle inequality. For the triangle inequality, we have in fact equality if and only if the summands have the same sign or at least one of them is zero. Therefore $=$ will occur in the first step if and only if $x+y-z$ and $x-y+z$ have the same sign (or one is zero) and also the other variants, thus in total if all of $x+y-z, x-y+z,-x+y+z$ are $ge0$ or all are $\le 0$. Thus we need $x+y\ge z, x+z\ge y, y+z\ge x$ or $x+y\le z, x+z\le y, y+z\le x.$

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    you're brilliant. thanks.2012-11-22