How to solve Rudin Chapter 2 problem 11? I have finished the entire problem except the 5th part. Roughly speaking the questions asks to show that over $\mathbb{R}$ prove/disprove that the distance function 
$$ d_5(x,y) = \frac{|x-y|}{1 + |x-y|} $$
Is a metric. I have proven symmetry and I have proven that if $d_5(x,y) = 0$ that $x = y$ so all that remains to be shown is the triangle inequality namely :
$$ \frac{|x-y|}{1 + |x-y|}+ \frac{|y-z|}{1 + |y-z|} \ge \frac{|x-z|}{1 + |x-z|} $$
This of course can be cast more simply in terms of $u,v$ in $\mathbb{R}$ as
$$ \frac{|u|}{1 + |u|}+ \frac{|v|}{1 + |v|} \ge \frac{|u+v|}{1 + |u+v|} $$
This reminds me an awful lot of contest style inequalities but i can't for the love of god place which one this is (not that it should matter, I should ideally be able to just generate a proof with ease). 
My one angle of attack was to do case work on fixing $u,v$ to be both positive or both negative, or a mix of positive and negative. But is there a easier way to proceed than that?

\begin{align*}
\dfrac{|u+v|}{1+|u+v|}&=1-\dfrac{1}{1+|u+v|}\\
&\leq1-\dfrac{1}{1+|u|+|v|}\\
&=\dfrac{|u|+|v|}{1+|u|+|v|}\\
&=\dfrac{|u|}{1+|u|+|v|}+\dfrac{|v|}{1+|u|+|v|}\\
&\leq\dfrac{|u|}{1+|u|}+\dfrac{|v|}{1+|v|}.
\end{align*}
