## Ted.knows.it Group Title Prove the following inequality: equation attached. one year ago one year ago

1. Ted.knows.it Group Title

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

2. Ted.knows.it Group Title

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

3. Ted.knows.it Group Title

@amistre64 ?

4. Ted.knows.it Group Title

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

5. Ted.knows.it Group Title

@satellite73 ?

6. klimenkov Group Title

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

7. Ted.knows.it Group Title

8. Ted.knows.it Group Title

no, they say nothing about using it.

9. klimenkov Group Title

$\ln \frac{ x_{1} + x_{2} + x_{3} + ... + x_{n} - n }{ x_{1} + x_{2} + x_{3} + ... + x_{n} }=\ln\frac{\frac{\sum_{i=1}^n(x_i-1)}{n}}{\frac{\sum_{i=1}^n x_i}{n}}=\ln\frac{\sum_{i=1}^n(x_i-1)}{n}-\ln\frac{\sum_{i=1}^n x_i}{n}$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:$\frac{\sum_{i=1}^n x_i}{n}\geqslant \sqrt[n]{\prod_{i=1}^n x_i} .$

10. klimenkov Group Title

Hm.. This is not very useful.

11. Ted.knows.it Group Title

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.

12. Ted.knows.it Group Title

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

13. klimenkov Group Title

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?

14. mukushla Group Title

one question...i want make myself sure :) $x_i>1$?

15. klimenkov Group Title

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

16. Ted.knows.it Group Title

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

17. Ted.knows.it Group Title

@mukushla , yes it is given by the function domain

18. mukushla Group Title

thank u :)

19. klimenkov Group Title

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.

20. Ted.knows.it Group Title

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

21. klimenkov Group Title

Yes. I wish you luck while proving this one.

22. Ted.knows.it Group Title

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