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

1. Ted.knows.it

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

2. Ted.knows.it

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

@amistre64 ?

4. Ted.knows.it

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

@satellite73 ?

6. klimenkov

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

7. Ted.knows.it

8. Ted.knows.it

no, they say nothing about using it.

9. klimenkov

$\ln \frac{ x_{1} + x_{2} + x_{3} + ... + x_{n} - n }{ x_{1} + x_{2} + x_{3} + ... + x_{n} }=\ln\frac{\frac{\sum_{i=1}^n(x_i-1)}{n}}{\frac{\sum_{i=1}^n x_i}{n}}=\ln\frac{\sum_{i=1}^n(x_i-1)}{n}-\ln\frac{\sum_{i=1}^n x_i}{n}$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

Hm.. This is not very useful.

11. Ted.knows.it

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

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

13. klimenkov

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

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

15. klimenkov

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

16. Ted.knows.it

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

17. Ted.knows.it

@mukushla , yes it is given by the function domain

18. mukushla

thank u :)

19. klimenkov

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

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

21. klimenkov

Yes. I wish you luck while proving this one.

22. Ted.knows.it

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