Question

Prove the following inequality: equation attached.

Answer 1

\[\ln \frac{ x_{1} + x_{2} + x_{3} + ... + x_{n} - n }{ x_{1} + x_{2} + x_{3} + ... + x_{n} } \ge \frac{ \ln ((x_{1}-1)(x_{2}-1)...(x_{n}-1)) - \ln(x_{1} x_{2} .... x_{2}) }{ n }\]

Answer 2

It is the point c) of a math analysis problem which gives the function: \[f : (1, \infty) \rightarrow \mathbb{R} \] \[f(x) = \ln \frac{ x-1 }{ x }\]

Answer 3

@amistre64 ?

Answer 4

I managed to bring it to the following form: \[(\frac{ x_{1}-1+x_{2}-1+...+x_{n}-1 }{ x_{1}+x_{2}+...+x_{n} } )^{n} \ge \frac{ (x_{1}-1)(x_{2}-1)...(x_{n}-1) }{ x_{1} x_{2} ... x_{n}}\]

Answer 5

@satellite73 ?

Answer 6

In your proff you are obliged to use this function?\[f(x) = \ln \frac{ x-1 }{ x }\]

Answer 7

sorry for the late answer.

Answer 8

no, they say nothing about using it.

Answer 9

\[\ln \frac{ x_{1} + x_{2} + x_{3} + ... + x_{n} - n }{ x_{1} + x_{2} + x_{3} + ... + x_{n} }=\ln\frac{\frac{\sum_{i=1}^n(x_i-1)}{n}}{\frac{\sum_{i=1}^n x_i}{n}}=\ln\frac{\sum_{i=1}^n(x_i-1)}{n}-\ln\frac{\sum_{i=1}^n x_i}{n}\]Analogically:\[\frac{ \ln ((x_{1}-1)(x_{2}-1)...(x_{n}-1)) - \ln(x_{1} x_{2} .... x_{2}) }{ n }=\ln\sqrt[n]{\prod_{i=1}^n(x_i-1)}-\ln\sqrt[n]{\prod_{i=1}^n x_i}\]Now use somehow that the arithmetic mean is bigger than the geometric mean:\[\frac{\sum_{i=1}^n x_i}{n}\geqslant \sqrt[n]{\prod_{i=1}^n x_i} .\]

Answer 10

Hm.. This is not very useful.

Answer 11

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.

Answer 12

however, i think that the mathematical induction is not the most viable solution in this case.

Answer 13

Ok. What is next?
\[\ln\frac{x_1+x_2-2}{x_1+x_2}\geqslant\frac{\ln\left((x_1-1)(x_2-1)\right)-\ln x_1x_2}{2}\]Are you able to prove it?

Answer 14

one question...i want make myself sure :) \[x_i>1\]?

Answer 15

I have a way to prove this using the inequality for arithmetic and geometric means.

Answer 16

now i was looking at that idea. it seems the right one...

Answer 17

@mukushla , yes it is given by the function domain

Answer 18

thank u :)

Answer 19

And also you need to know that \(f(x)=\ln\frac{x-1}{x}\) is an increasing function. Think about the idea I wrote above. It is not so difficult.

Answer 20

yeah, so all in all, the problem goes down to proving that arithmetic mean is bigger than the geometric mean..

Answer 21

Yes. I wish you luck while proving this one.

Answer 22

thanks for your time. :)

Answer 23

this one from me...i hope it will help\[f''(x)=\frac{-2x+1}{x^2(x-1)^2}<0 \ \ , \ \ \forall x>1\]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