Inegalitate logaritmi.

inc.nci Vin Oct 19, 2012 7:37 pm · **inc.nci** Mesaje : 15 Reputatie : 0 Data de inscriere : 11/10/2012

Inegalitate logaritmi.

inc.nci Vin Oct 19, 2012 7:37 pm

$a,b,c\epsilon (1,\infty )Demonstrati \log_{a}(b^{\log_{a}b})+\log_{b}(c^{\log_{b}c})+\log_{c}(a^{\log_{c}a})\geq \log_{b}a+\log_{c}b+\log_{a}c$

**Admin** Mier Noi 07, 2012 6:47 pm

Re: Inegalitate logaritmi.

Admin Mier Noi 07, 2012 6:47 pm

$\begin{array}{l} \left\{ \begin{array}{l} {\log _a}{b^{{{\log }_a}b}} = \log _a^2b \to {\text{idem restul}}{\rm{.}}\\ {\rm{lo}}{{\rm{g}}_b}a = \frac{1}{{{{\log }_a}b}}\\ {\rm{notam:}}{\log _a}b = x \to {\text{idem restul}}{\rm{. }} \end{array} \right.\\ \Rightarrow {x^2} + {y^2} + {z^2} \ge \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \Leftrightarrow xyz\left( {{x^2} + {y^2} + {z^2}} \right) \ge xy + yz + zx.\\ \left\{ \begin{array}{l} \bullet a,b,c \ge 1 \Rightarrow \prod {{{\log }_a}b} = xyz = 1\\ \bullet \left( {{x^2} + {y^2} + {z^2}} \right) \ge xy + yz + zx \end{array} \right. \Rightarrow xyz\left( {{x^2} + {y^2} + {z^2}} \right) \ge xy + yz + zx \end{array}$