Prove $\frac{|a+b|}{1+|a+b|}<\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}$. Prove that for every $a, b\in R\setminus\{0\}$ is correct this inequality:
$$\frac{|a+b|}{1+|a+b|}<\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}.$$

Function 
$f(x)=\frac{x}{1+x}$,   $(x>0)$
is breeders because
$f'(x)=\frac{1}{(1+x)^2}>0$,
therefore by
$|a+b|<|a|+|b|$
have:
$f(|a+b|)<f(|a|+|b|)$
$\frac{|a+b|}{1+|a+b|}<\frac{|a|+|b|}{1+|a|+|b|}$=$\frac{|a|}{1+|a|+|b|}+\frac{|b|}{1+|a|+|b|}$
