# 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 monotonically increasing because

$$f'(x)=\frac{1}{(1+x)^2}>0$$,

therefore by

$$|a+b|<|a|+|b|$$

have:

$$f(|a+b|)

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

• I have never heard the term "breeders" before. Does it mean increasing? Commented Sep 11, 2012 at 19:55
• What is breeders? Commented Sep 11, 2012 at 19:56
• i would rather prefer the term "monotonically increasing". Commented Sep 11, 2012 at 19:58
• The last expression is not the RHS of the desired inequality.
– Did
Commented Sep 12, 2012 at 19:24

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 $m\neq 0 \neq n\neq 0 \neq p$.