# Metric space proof

1. Sep 16, 2010

### mynameisfunk

ok i am stumped on a proof i am trying to construct of a metric:
d(x,y)=$$\frac{|x-y|}{1+|x-y|}$$

so, out of the 3 requirements to be a metric, the first 2 are trivial and I am just working on proving the triangle inequality...

i need $$\frac{|x-y|}{1+|x-y|}$$ $$\leq$$ $$\frac{|x-z|}{1+|x-z|}$$ + $$\frac{|z-y|}{1+|z-y|}$$

p2(1+q+r+qr) $$\leq$$ q2(1+p+r+pr)+r2(1+p+q+pq)

can i now go to:
p(1+q+r+qr) $$\leq$$ q(1+p+r+pr)+r(1+p+q+pq) ???

2. Sep 18, 2010

### csopi

If you simplify the inequality you get:

p<=q+r+(some other positive expressions)

This is true, because p<=q+r by the properties of the absolute value.

3. Sep 18, 2010

### fluxions

hint: first show that
$$\frac{u}{1+u} \leq \frac{v}{1+v}$$
whenever
$$0 \leq u \leq v$$.

4. Sep 18, 2010

### snipez90

The method you're suggesting is clever, but the OP almost had a correct solution. He seems to have written down an unjustified inequality and then questioned the correct inequality (where the solution then immediate falls out by multiplying everything out and canceling).

5. Sep 18, 2010

### mynameisfunk

i dont know If i posted enough info but, I had set |x-y|= p , |x-z| = q , |z-r| = r , and since the triangle equality holds, proving p <= q + r will suffice

6. Sep 19, 2010

### snipez90

Defining your variables would have probably gotten you more responses, but it was pretty easy to figure out what p,q,r was. The only confusing part was that the last inequality you wrote in the original post is not a consequence of the inequality before it. The last inequality is what you should get upon clearing denominators. Then multiply out and you should get what csopi wrote. It's also clear that the steps are reversible.