Proving an interesting property of absolute values: $\frac{|x+y|}{1+|x+y|}\le\frac{|x|}{1+|y|}+\frac{|y|}{1+|y|}$ 
Question
If $x$ and $y$ are real numbers, show that
$$\frac{|x+y|}{1+|x+y|}≤\frac{|x|}{1+|y|}+\frac{|y|}{1+|y|}$$

I am having difficulty in proving this equation. I don't know where to start. Your help will be highly appreciated.

Consider the function 
$$
f(t)=\frac{t}{1+t}
$$
where $t \ge 0$. It is immediate to see that it is an increasing function on its domain. Now, since $|x+y| \le |x|+|y|$, you will have
$$
\frac{|x+y|}{1+|x+y|}\le \frac{|x|}{1+|x|+|y|}+\frac{|y|}{1+|x|+|y|}
$$
and, since $\frac{1}{1+|x|+|y|} \le \frac{1}{1+|x|}$, you can conclude.
P.S.
Observe that the above argument holds also for the inequality you write... in the comment, I write the classical inequality students must solve in their analysis courses!
