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.

We know that $|x+y|\leq|x|+|y|$. You need to prove $$\frac{|x+y|}{1+|x+y|}\leq\frac{|x|}{1+|x|}+\frac{|y|}{1+|y|}$$ $$or,~|x+y|.(1+|x|)(1+|y|)\leq(1+|x+y|)(|x|.(1+|y|)+|y|.(1+|x|))$$
by cross multiplication. By expanding we get
$$|x+y|\leq|x|+|y|+2|x||y|+|x||y||x+y|$$
which is obviously true since $2|x||y|+|x||y||x+y|\geq0$.
