Prove $\frac{|a+b|}{1+|a+b|}<\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}$.

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:

$f(|a+b|)<f(|a|+|b|)$

$\frac{|a+b|}{1+|a+b|}<\frac{|a|+|b|}{1+|a|+|b|}$=$\frac{|a|}{1+|a|+|b|}+\frac{|b|}{1+|a|+|b|}$

I. General case: Let $m,n$ and $p$ be positive real numbers and $m \leq n+p$. Let's prove that:

$$\frac{m}{1+m} \leq \frac{n}{1+n}+\frac{p}{1+p},$$

This inequality is equivalent to : $$m(1+m)(1+n) \leq n(1+p)(1+m)+p(1+n)(1+m) \Leftrightarrow$$ $$m+mn+mp \leq n+np+nm+p+pm+pn+mpn \Leftrightarrow$$ $$m \leq n+p+2np+mnp,$$ which it is true because $m \leq n+p$ and $m,n,p \in \mathbb{R_{+}}$.

II. The inequality: We know that $\|a+b\| \leq \|a\|+\|b\|$ and so we take: \begin{eqnarray} m&=&\|a+b\| \\ n&=&\|a\|\\ p&=&\|b\|, \end{eqnarray} because $\|a+b\|, \|a\|, \|b\| \in \mathbb{R_{+}}.$

The inequality is strict if $m\neq 0 \neq n\neq 0 \neq p$.