# Metric

• Jan 15th 2010, 02:38 AM
slevvio
Metric
Hello there, I have this question about metrics, but the first part asks me to prove this horrible identity which I am having difficulty with. I would appreciate if somebody could point me in the right direction before I cover all my writing supplies with algebra.

Let $\displaystyle a,b,c$ be non-negative real numbers such that $\displaystyle a \le b+c$. Show that $\displaystyle \frac{b}{1+b} + \frac{c}{1+c} - \frac{a}{1+a} \ge 0$.

Thanks again. (Happy)
• Jan 15th 2010, 04:47 AM
Laurent
Quote:

Originally Posted by slevvio
Hello there, I have this question about metrics, but the first part asks me to prove this horrible identity which I am having difficulty with. I would appreciate if somebody could point me in the right direction before I cover all my writing supplies with algebra.

Let $\displaystyle a,b,c$ be non-negative real numbers such that $\displaystyle a \le b+c$. Show that $\displaystyle \frac{b}{1+b} + \frac{c}{1+c} - \frac{a}{1+a} \ge 0$.

Thanks again. (Happy)

$\displaystyle f:x\mapsto\frac{x}{1+x}$ is increasing on $\displaystyle \mathbb{R}_+$, thus $\displaystyle f(a)\leq f(b+c)$, and the question becomes: $\displaystyle f(b+c)\leq f(b)+f(c)$? i.e. $\displaystyle \frac{b+c}{1+b+c}\leq \frac{b+c+2bc}{1+b+c+bc}$? If $\displaystyle \frac{u}{v}\leq \frac{x}{y}$, then $\displaystyle \frac{u}{v}\leq\frac{u+x}{v+y}$. With suitable $\displaystyle u,v,x,y$, this concludes.
• Jan 16th 2010, 01:51 AM
Shanks
Notice that the function Laurent has defined is increasing on [0, infinity).
thus f(b)+f(c)>=f(b+c)>=f(a) since b+c>=a.
• Jan 18th 2010, 08:11 AM
slevvio
Thanks very much but how do i show that http://www.mathhelpforum.com/math-he...9c98790e-1.gif ?
• Jan 18th 2010, 09:22 AM
Plato
Lemma:
$\displaystyle \begin{gathered} 0 \leqslant x \leqslant y \Rightarrow \quad x+xy \leqslant y + xy \hfill \\ \Rightarrow \quad x(1 + y) \leqslant y(1 + x) \Rightarrow \quad \frac{x} {{1 + x}} \leqslant \frac{y}{{1 + y}} \hfill \\ \end{gathered}$

Using that lemma, we know that $\displaystyle a\le b+c$ so

$\displaystyle \frac{a}{1 +a } \leqslant \frac{b+c}{1+b+c} \leqslant \frac{b}{1+b}+ \frac{c}{1+c}$
• Jan 19th 2010, 07:46 AM
Laurent
nb: The fact that $\displaystyle x\mapsto \frac{x}{1+x}=1-\frac{1}{1+x}$ is increasing is almost obvious with second writing (composition of decreasing maps => increasing)
Quote:

Originally Posted by slevvio
Thanks very much but how do i show that http://www.mathhelpforum.com/math-he...9c98790e-1.gif ?

Just multiply by v(v+y) (i.e. get the denominators away) and simplify. But this step is somewhat direct (btw it corresponds to the last inequality in Plato's post, that comes with no further argument), my trick was just a way to avoid any computation by reducing to an elementary inequality.