I am pretty sure that this is such a stupid, stupid question, but how do you prove that $d_2(x,y)={|x-y|\over {1+|x-y|}}$ satisfying the third condition to be a metric, which is the triangle inequality.

I know that I should really be able to do this, but I just can't, I keep seeing a contradiction. Please help me.


$d_2 (x,y)+d_2 (y,z)=\frac{d(x,y)}{1+d(x,y)}+\frac{d(y,z)}{1+d(y,z)}$

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

$= 1-\frac{1}{1+d(x,y)+d(y,z)} \geq 1-\frac{1}{1+d(x,z)}$

$= \frac{d(x,z)}{1+d(x,z)}=d_2 (x,z)$


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