Distance function is in fact a metric I know I should be able to show this, but for some reason I am having trouble. I need to show that $$d(x,y) = \frac{|x-y|}{1+|x-y|}$$ is a metric on $\Bbb R$ where $|*|$ is the absolute value metric. I am getting confused trying to show that the triangle inequality holds for this function. My friend also said that he proved that this distance function defines a metric even if you replace $|*|$ with any other metric. So I'd like to try and show both, but I cannot even get the specific case down first. Please help.

Hint:
Put $f(t) = \frac{t}{1+t} $. Verify yourself that 
$$ f'(t) = \frac{1}{(1+t)^2 } $$
Hence, $f'(t) \geq 0 $ for all $t$. In particular $f$ is an increasing function. In other words, we have 
$$ |x+y| \leq |x| + |y| \implies f(|x+y|) \leq f(|x|+|y|) \implies .....$$
