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

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

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

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

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

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

klimenkovBest 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:\[\frac{\sum_{i=1}^n x_i}{n}\geqslant \sqrt[n]{\prod_{i=1}^n x_i} .\]
 one year ago

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

Ted.knows.itBest ResponseYou've already chosen the best response.0
i 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.
 one year ago

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

klimenkovBest ResponseYou've already chosen the best response.1
Ok. 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?
 one year ago

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

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

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

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

klimenkovBest ResponseYou've already chosen the best response.1
And 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.
 one year ago

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

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

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

mukushlaBest ResponseYou've already chosen the best response.1
this 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
 one year ago
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