# proving a metric

I'm trying to show that given a metric $d(x,y)$ show that $d_0(x,y) =\frac{d(x,y)}{1+d(x,y)}$ is also a metric..

It's trivial to show the first two properties, that is, $d_0\geq 0$ & for $x=y \Longleftrightarrow$ $d(x,y) =0$ and that $d_0(x,y) = d_0(y,x)$.

However, how would I go about showing that $d_0(x,y) \leq d_0(x,z) +d_0(z,y)$. This is a bit tricky for me.

Here's what I've done so far.

$$d_0(x,y) = \frac{d(x,y)}{1+d(x,y)}\leq \frac{d(x,z)+d(z,y)}{1+d(x,z)+d(z,y)}=\frac{d(x,z)}{1+d(x,z)+d(z,y)}+\frac{d(z,y)}{1+d(x,z)+d(z,y)}$$

I'm not sure where to proceed from here.

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Well, show us what you've tried and where you're getting hung up; it will make it much easier for us to help you along. –  Nick Peterson Mar 30 '14 at 19:52
possible duplicate of math.stackexchange.com/questions/686792/… –  Sujaan Kunalan Mar 30 '14 at 19:57

As $d(x,y), d(y,z) \geq 0$ we have that $$1 + d(x,y) + d(y,z) \geq 1 + d(x,y) > 0$$So we then get that $$\frac{d(x,y)} {1 + d(x,y) + d(y,z) } \leq \frac{d(x,y)} {1 + d(x,y)} = d_0(x,y)$$ And we then get $$d_0 (x,z) \leq \frac{d(x,y)} {1 + d(x,y) + d(y,z) } + \frac{d(y,z)} {1 + d(x,y) + d(y,z) } \leq \frac{d(x,y)} {1 + d(x,y)} + \frac{d(y,z)} {1 + d(y,z)}$$ $$= d_0(x,y) + d_0 (y,z)$$

So we have the triangle inequality, so putting what you already proved together with this, we have that $d_0$ is a metric.

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Consider the function $f(t)=\frac{t}{1+t}$, this is a monotonic increasing function, since $f'(t)=\frac{1}{(t+1)^2}\ge0, \forall t\in \mathbb{R}.$

Thus for $s$ positive $f(t)\leq f(t+s)$, now apply the triangle inequality to get your desired result.

So we get $d_0(x,y)=\frac{|x-y|}{1+|x-y|}\leq \frac{|x-z|+|z-y|}{1+|x-z|+|z-y|}=\frac{|x-z|}{1+|x-z|+|z-y|}+\frac{|z-y|}{1+|x-z|+|z-y|}\leq \frac{|x-z|}{1+|x-z|}+\frac{|z-y|}{1+|z-y|}$

=$d_0(x,z)+d_0(z,y)$

Thus $d_0(x,y)\leq d_0(x,z)+d_0(z,y)$.

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