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Ted.knows.it
 one year ago
Best ResponseYou've already chosen the best response.0\[\ln \frac{ x_{1} + x_{2} + x_{3} + ... + x_{n}  n }{ x_{1} + x_{2} + x_{3} + ... + x_{n} } \ge \frac{ \ln ((x_{1}1)(x_{2}1)...(x_{n}1))  \ln(x_{1} x_{2} .... x_{2}) }{ n }\]

Ted.knows.it
 one year ago
Best ResponseYou've already chosen the best response.0It is the point c) of a math analysis problem which gives the function: \[f : (1, \infty) \rightarrow \mathbb{R} \] \[f(x) = \ln \frac{ x1 }{ x }\]

Ted.knows.it
 one year ago
Best ResponseYou've already chosen the best response.0I managed to bring it to the following form: \[(\frac{ x_{1}1+x_{2}1+...+x_{n}1 }{ x_{1}+x_{2}+...+x_{n} } )^{n} \ge \frac{ (x_{1}1)(x_{2}1)...(x_{n}1) }{ x_{1} x_{2} ... x_{n}}\]

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.1In your proff you are obliged to use this function?\[f(x) = \ln \frac{ x1 }{ x }\]

Ted.knows.it
 one year ago
Best ResponseYou've already chosen the best response.0sorry for the late answer.

Ted.knows.it
 one year ago
Best ResponseYou've already chosen the best response.0no, they say nothing about using it.

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.1\[\ln \frac{ x_{1} + x_{2} + x_{3} + ... + x_{n}  n }{ x_{1} + x_{2} + x_{3} + ... + x_{n} }=\ln\frac{\frac{\sum_{i=1}^n(x_i1)}{n}}{\frac{\sum_{i=1}^n x_i}{n}}=\ln\frac{\sum_{i=1}^n(x_i1)}{n}\ln\frac{\sum_{i=1}^n x_i}{n}\]Analogically:\[\frac{ \ln ((x_{1}1)(x_{2}1)...(x_{n}1))  \ln(x_{1} x_{2} .... x_{2}) }{ n }=\ln\sqrt[n]{\prod_{i=1}^n(x_i1)}\ln\sqrt[n]{\prod_{i=1}^n x_i}\]Now use somehow that the arithmetic mean is bigger than the geometric mean:\[\frac{\sum_{i=1}^n x_i}{n}\geqslant \sqrt[n]{\prod_{i=1}^n x_i} .\]

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.1Hm.. This is not very useful.

Ted.knows.it
 one year ago
Best ResponseYou've already chosen the best response.0i tried to analyse the problem step by step and started with x1 \[\ln \frac{ x_{1}  1 }{ x_{1} } \ge \frac{ \ln (x_{1}  1)  \ln x_{1} }{ 1 }\] which is true.

Ted.knows.it
 one year ago
Best ResponseYou've already chosen the best response.0however, i think that the mathematical induction is not the most viable solution in this case.

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.1Ok. What is next? \[\ln\frac{x_1+x_22}{x_1+x_2}\geqslant\frac{\ln\left((x_11)(x_21)\right)\ln x_1x_2}{2}\]Are you able to prove it?

mukushla
 one year ago
Best ResponseYou've already chosen the best response.1one question...i want make myself sure :) \[x_i>1\]?

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.1I have a way to prove this using the inequality for arithmetic and geometric means.

Ted.knows.it
 one year ago
Best ResponseYou've already chosen the best response.0now i was looking at that idea. it seems the right one...

Ted.knows.it
 one year ago
Best ResponseYou've already chosen the best response.0@mukushla , yes it is given by the function domain

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.1And also you need to know that \(f(x)=\ln\frac{x1}{x}\) is an increasing function. Think about the idea I wrote above. It is not so difficult.

Ted.knows.it
 one year ago
Best ResponseYou've already chosen the best response.0yeah, so all in all, the problem goes down to proving that arithmetic mean is bigger than the geometric mean..

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.1Yes. I wish you luck while proving this one.

Ted.knows.it
 one year ago
Best ResponseYou've already chosen the best response.0thanks for your time. :)

mukushla
 one year ago
Best ResponseYou've already chosen the best response.1this one from me...i hope it will help\[f''(x)=\frac{2x+1}{x^2(x1)^2}<0 \ \ , \ \ \forall x>1\]so \(f\) is concave. Then using the Jensen’s inequality we obtain:\[\ln(1\frac{1}{\frac{x_1+x_2+...+x_n}{n}}) \ge \frac{1}{n} \sum_{i=1}^{n} \ln(1\frac{1}{x_i})\]we are done
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