Home/ Intrebari/Q 82431
Urmator
In Process

AniDeȘcoală.ro Latest Intrebari

viperaza
Pe: 2013-08-05T14:15:43+03:00 In: In:

inecuatie logaritmi(2)

OLM Brasov 2013
Dacă a,b,c>1 demonstrati inegalitatea:

    \[log_{a}\frac{a^2+b^{2}+c^{2}}{a+b+c}+log_{b}\frac{a^2+b^{2}+c^{2}}{a+b+c}+log_{c}\frac{a^2+b^{2}+c^{2}}{a+b+c}\geq 3\]

Similare

4 raspunsuri

  1. gunty
    maestru (V)

        \[ \begin{array}{l} Consideram\;functia\;convexa\;f]{{abc}} \le \frac{{a + b + c}}{3}_{\left( 2 \right)} \\ \left( 1 \right),\left( 2 \right) \Rightarrow \sqrt[3]{{abc}} \le \frac{{a^2 + b^2 + c^2 }}{{a + b + c}} \Rightarrow \\ \Rightarrow \log _a \frac{{a^2 + b^2 + c^2 }}{{a + b + c}} + \log _b \frac{{a^2 + b^2 + c^2 }}{{a + b + c}} + \log _c \frac{{a^2 + b^2 + c^2 }}{{a + b + c}} \ge \log _a \sqrt[3]{{abc}} + \log _b \sqrt[3]{{abc}} + \log _c \sqrt[3]{{abc}} \\ Deci\;e\;destul\;sa\;aratam\;ca: \\ \log _a \sqrt[3]{{abc}} + \log _b \sqrt[3]{{abc}} + \log _c \sqrt[3]{{abc}} \ge 3 \\ \Leftrightarrow \frac{1}{3}\left( {3 + \log _a b + \log _a b + \log _b a + \log _b c + \log _c a + \log _c b} \right) \ge 3 \\ \Leftrightarrow \left( {\log _a b + \log _b a} \right) + \left( {\log _b c + \log _c b} \right) + \left( {\log _c a + \log _a c} \right) \ge 6 \\ \Leftrightarrow \left( {\log _a b + \frac{1}{{\log _a b}}} \right) + \left( {\log _b c + \frac{1}{{\log _b c}}} \right) + \left( {\log _c a + \frac{1}{{\log _c a}}} \right) \ge 6_{\left( {adev.} \right)} \\ \end{array} \]

  2. viperaza

    Multumesc! Am stat mult pe ea, nu m-am gandit la rezolvarea asta.

  4. Green eyes
    maestru (V)

    Salut,

  5. TheodorMunteanu

    demonstrati ca \sum_{i=1}^n \log_{a_i} \frac{\sum_{i=1}^n a_i^2}{\sum_{i=1}^n a_i}\geq noricare ar fi a_i>1

    Rez:din inegalitatea C-B-S,\frac{\sum_{i=1}^n a_i^2}{\sum_{i=1}^n a_i}\geq \frac{\sum_{i=1}^n a_i}{n}\geq \sqrt[n]{a_1a_2...a_n}(din ineg.mediilor)
    apoi,\sum_{i=1}^n \log_{a_i}\sqrt[n]{a_1a_2...a_n}=\sum_{i=1} \frac{ln \sqrt[n]{a_1a_2...a_n}}{ln a_i}=\sum_{i=1}^n \frac{ln a_1+ln a_2+...+ln a_n}{n}\cdot \left(\frac{1}{ln a_1}+\frac{1}{ln a_2}+...+\frac{1}{ln a_n}\right)care din nou din inegalitatea mediilor e mai mare sau egala decat n^2/n=n

    PS:am folosit binecunoscuta inegalitatea \left(x_1+...+x_n\right)\left(\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}\right)\geq n^2

Raspunde

Raspunde