Prove that for every $a, b\in R\setminus\{0\}$ is correct this inequality:




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

is breeders because


therefore by





  • $\begingroup$ I have never heard the term "breeders" before. Does it mean increasing? $\endgroup$ – Alex Becker Sep 11 '12 at 19:55
  • $\begingroup$ What is breeders? $\endgroup$ – Euler....IS_ALIVE Sep 11 '12 at 19:56
  • 2
    $\begingroup$ i would rather prefer the term "monotonically increasing". $\endgroup$ – mythealias Sep 11 '12 at 19:58
  • $\begingroup$ The last expression is not the RHS of the desired inequality. $\endgroup$ – Did Sep 12 '12 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 $m\neq 0 \neq n\neq 0 \neq p$.


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