TITLE: Prove $\frac{|a+b|}{1+|a+b|}<\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}$.
QUESTION [5 upvotes]: Prove that for every $a, b\in R\setminus\{0\}$ is correct this inequality:
$$\frac{|a+b|}{1+|a+b|}<\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}.$$

REPLY [2 votes]: 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 $m\neq 0 \neq n\neq 0 \neq p$.