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Can the inequality

$2y\ge|x-y|+|y-z|-|x-z|\quad\, \forall x,y,z\geq0$

be simplified? It looks similar to the triangle inequality.

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    I have simplified it, in a form that brings out the geometry. It turns out that one *can* find an inequality which is equivalent to the given one. Inequalities that are not always true can still be useful within their range of validity.2012-02-16

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I will assume that the question is to be interpreted as follows.

Consider the inequality $2y\ge|x-y|+|y-z|-|x-z|$. Can this be replaced by a simpler inequality? We are allowed to assume that $x$, $y$, and $z$ are non-negative.

Certainly the above inequality fails sometimes, for example when $y=0$ and $x=z=1$. So let us explore under what conditions it holds.

A sensible approach is to draw a picture, actually three pictures. By symmetry we may temporarily assume that $x\le z$. So there are three possible positions for $y$. It can be (i) greater than $z$ or (ii) between $x$ and $z$ or (iii) less than $x$.

A little play shows that the inequality holds in cases (i) and (ii). Case (iii) is more interesting, so we do the details. A cursory look at the picture shows that in case (iii), $|x-y| + |y-z|= (x-y) +(z-y) = z-x +2(x-y).$ Since $|z-x|=z-x$, our inequality becomes $2y \ge 2(x-y)$, or equivalently $2y\ge x$.

We only let $x\le z$ for the sake of the picture, so it is time to symmetrize. We conclude that the inequality holds precisely if $\frac{1}{2}\min(x,z)\le y \le \min(x,z).$

I hope that this simplification is a useful one.

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Do you have some constraints about $x, y$ and $z$ ? Because for instance, if you take $y=-1$ and $x=z=0$, your inequality does not hold.

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    Yeah, I forgot to say that x,y,z are non negative.2012-02-15
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Well, using triangle inequality, $|x-z|\leq|x-y|+|y-z|.$ This implies the right hand side is always bigger than zero. So $2y\geq0$.

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    I would like to preserve the variables x,y,z.2012-02-15