Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
I am puzzled. Let $y=0$, $x=z=1$. Then the inequality $2y\ge|x-y|+|y-z|-|x-z|$ seems to say that $0\ge 2$. Or are you trying to find a simpler-looking inequality which is equivalent to the one you gave, and thus fails sometimes? –  André Nicolas Feb 16 '12 at 0:24
    
@AndréNicolas I was originally trying to find an inequality equivalent to the given one. However, since this inequality is not always true, I guess it is not really necessary to try to simplify it. –  MathMajor Feb 16 '12 at 1:14
    
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. –  André Nicolas Feb 16 '12 at 1:30

3 Answers 3

up vote 6 down vote accepted

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.

share|improve this answer

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.

share|improve this answer
    
Yeah, I forgot to say that x,y,z are non negative. –  MathMajor Feb 15 '12 at 21:14

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

share|improve this answer
    
I would like to preserve the variables x,y,z. –  MathMajor Feb 15 '12 at 22:31

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.