# Simple Metric Question

1. Jan 17, 2010

### insynC

1. The problem statement, all variables and given/known data

Let X be a metric space with metric d. Show that the space X, where distance is measured by d' = d/(1+d), is also a metric space.

2. Relevant equations

Three requirements of a distance function for X to be a metric space:
1. d(x,y) = 0 <=> x=y
2. d(y,x) = d(x,y)
3. d(x,z) <= d(x,y) + d(y,z)

3. The attempt at a solution

It's demonstrating the triangle inequality (3. above) that has me stumped.

My starting point is to try and use (1) d(x,z) <= d(x,y) + d(y,z) to demonstrate (2) d'(x,z) <= d'(x,y) + d'(y,z).

I tried manipulating the RHS of (2) to get a common denominator and then tried to 'bash' it out, without any success. I then tried to manipulate the inequalities, but I can't get (2) whilst maintaining <= as opposed to simply <.

The problem wasn't set as a difficult one, so I presume there is something simple that I can't seem to see.

Any help would be greatly appreciated!

Cheers

2. Jan 17, 2010

### snipez90

Hmmm your first approach should work (I remember it worked for me). Define $f(t) = \frac{1}{1+t}.$ You basically need to show $f(a) + f(b) \geq f(c)$ given $a + b \geq c \geq 0.$ Multiplying out the first inequality gives
$$a + ac + ab + abc + b + ba + bc + abc \geq c + ca + cb + abc,$$
and yeah this should be true unless I made a mistake somewhere.

3. Jan 17, 2010

### insynC

Thanks for the help, I think that sorts it.