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.

Suppose that $d(\cdot,\cdot)$ is a arbitrary metric and $\bar{d}(x,y)=\frac{d(x,y)}{1+d(x,y)}$, then we prove the triangle inequality for $\bar{d}(\cdot,\cdot)$ as following:

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

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

$$ |x+y| \leq |x| + |y| \implies f(|x+y|) \leq f(|x|+|y|) \implies .....$$