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