1. ## Problem with inequality

Hi!, I have problems with this Proposition

PROPOSITION

Let $d,d^{\prime}$ and $d^{\prime\prime}$ metric space. For any $x,y \in{\mathbb{R^n}}$ we have

$d(x,y)\leq{d^{\prime}(x,y)}\leq{nd^{\prime\prime}( x,y)}$ , where $d(x,y)= [ \displaystyle\sum_{i=1}^N (x_i-y_i)^2]^{1/2}$ ,

$d^{\prime}(x,y)= \displaystyle\sum_{i=1}^N |x_i-y_i|$ , $max_{1\leq{i}\leq{n}} |x_i-y_i|$

I have problems with proof the inequalities

2. ## Re: Problem with inequality

Originally Posted by cristianoceli
Hi!, I have problems with this Proposition

PROPOSITION

Let $d,d^{\prime}$ and $d^{\prime\prime}$ metric space. For any $x,y \in{\mathbb{R^n}}$ we have

$d(x,y)\leq{d^{\prime}(x,y)}\leq{nd^{\prime\prime} (x,y)}$ , where $d(x,y)= [ \displaystyle\sum_{i=1}^N (x_i-y_i)^2]^{1/2}$ ,

$d^{\prime}(x,y)= \displaystyle\sum_{i=1}^N |x_i-y_i|$ , $max_{1\leq{i}\leq{n}} |x_i-y_i|$

I have problems with proof the inequalities
I have replaced "tex" and "/tex" with dollar signs.

3. ## Re: Problem with inequality

It is a geometric property of triangles that the sum of the lengths of two sides is always greater than the third. So, for 2 dimensions, the inequality follows immediately from the fact that d(x,y) is the length of the hypotenuse while d'(x,y) is either that exact same length or it is the sum of the lengths of the other two sides of the triangle. For higher dimensions, you would need a generalization of that fact.

http://www.cs.bc.edu/~alvarez/NDPyt.pdf

This gives an idea of how to start.

Thanks!!!!