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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?




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

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

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How did you prove $\ge$ in the first place? A proof for when equality holds usually emerges by carefully looking at each step. – Hagen von Eitzen Nov 22 '12 at 18:37
@HagenvonEitzen Well I multiplied with 2 and used the triangle inequation – sheldoor Nov 22 '12 at 18:38
I assume that you used some version of the triangle inequality, applied two or three times, to get the result. Look at the equality cases in each of the inequalities that you brought together. – Fly by Night Nov 22 '12 at 18:38
For the triangle inequality you may use $|a+b|\le |a|+|b|$ with equality iff $a,b$ don't have different signs (that is iff $a,b\ge0$ or $a,b\le 0$). – Hagen von Eitzen Nov 22 '12 at 18:41
@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? – sheldoor Nov 22 '12 at 18:43
up vote 1 down vote accepted

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. – sheldoor Nov 22 '12 at 18:50

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