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Ted.knows.it
 2 years ago
Prove the following inequality: equation attached.
Ted.knows.it
 2 years ago
Prove the following inequality: equation attached.

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

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

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

Ted.knows.it
 2 years 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
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1one question...i want make myself sure :) \[x_i>1\]?

klimenkov
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0now i was looking at that idea. it seems the right one...

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

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

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

mukushla
 2 years 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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