Inegalitati logaritmice

elena123 Joi Dec 06, 2012 1:55 am

Demonstreaza inegalitatile:

log n(n+1)>log n+1 (n+2) oricare n e N*, n ≥2

log a+1 b < log a ba (b+1) , a,b >1

Titlul editat de Unu

**Admin** Joi Dec 06, 2012 9:09 pm

log n(n+1)>log n+1 (n+2) oricare n e N*, n ≥2

Putem scrie logaritmul $log_{n}{n+1}=\frac{1}{log_{n+1}{n}}$ astfel, inegalitatea se poate scrie ca find :
$\frac{1}{{lo{g_{n + 1}}n}} > lo{g_{n + 1}}\left( {n + 2} \right) \Leftrightarrow 1 > lo{g_{n + 1}}n \times lo{g_{n + 1}}\left( {n + 2} \right)$
Acum, din (MA-MG)₂ avem:
$\begin{array}{l} \frac{1}{{lo{g_{n + 1}}n}} > lo{g_{n + 1}}\left( {n + 2} \right) \Leftrightarrow \\ lo{g_{n + 1}}n \times lo{g_{n + 1}}\left( {n + 2} \right) < \frac{{{{\log }_{n + 1}}n + {{\log }_{n + 1}}\left( {n + 2} \right)}}{2} = \frac{{{{\log }_{n + 1}}n\left( {n + 2} \right)}}{2} < \frac{{{{\log }_{n + 1}}{{\left( {n + 1} \right)}^2}}}{2} = 1 \end{array}$
Care este exact ceea ce urmăream !

Inegalitati logaritmice

Inegalitati logaritmice

Re: Inegalitati logaritmice