Inequality Similar to Triangle Inequality 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.
 A: 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.
A: 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.
A: Well, using triangle inequality, $|x-z|\leq|x-y|+|y-z|.$ This implies the right hand side is always bigger than zero. So $2y\geq0$.
