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Ted.knows.it Group Title

Prove the following inequality: equation attached.

  • one year ago
  • one year ago

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  1. Ted.knows.it Group Title
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    \[\ln \frac{ x_{1} + x_{2} + x_{3} + ... + x_{n} - n }{ x_{1} + x_{2} + x_{3} + ... + x_{n} } \ge \frac{ \ln ((x_{1}-1)(x_{2}-1)...(x_{n}-1)) - \ln(x_{1} x_{2} .... x_{2}) }{ n }\]

    • one year ago
  2. Ted.knows.it Group Title
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    It is the point c) of a math analysis problem which gives the function: \[f : (1, \infty) \rightarrow \mathbb{R} \] \[f(x) = \ln \frac{ x-1 }{ x }\]

    • one year ago
  3. Ted.knows.it Group Title
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    @amistre64 ?

    • one year ago
  4. Ted.knows.it Group Title
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    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
  5. Ted.knows.it Group Title
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    @satellite73 ?

    • one year ago
  6. klimenkov Group Title
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    In your proff you are obliged to use this function?\[f(x) = \ln \frac{ x-1 }{ x }\]

    • one year ago
  7. Ted.knows.it Group Title
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    sorry for the late answer.

    • one year ago
  8. Ted.knows.it Group Title
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    no, they say nothing about using it.

    • one year ago
  9. klimenkov Group Title
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    \[\ln \frac{ x_{1} + x_{2} + x_{3} + ... + x_{n} - n }{ x_{1} + x_{2} + x_{3} + ... + x_{n} }=\ln\frac{\frac{\sum_{i=1}^n(x_i-1)}{n}}{\frac{\sum_{i=1}^n x_i}{n}}=\ln\frac{\sum_{i=1}^n(x_i-1)}{n}-\ln\frac{\sum_{i=1}^n x_i}{n}\]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:\[\frac{\sum_{i=1}^n x_i}{n}\geqslant \sqrt[n]{\prod_{i=1}^n x_i} .\]

    • one year ago
  10. klimenkov Group Title
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    Hm.. This is not very useful.

    • one year ago
  11. Ted.knows.it Group Title
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    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
  12. Ted.knows.it Group Title
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    however, i think that the mathematical induction is not the most viable solution in this case.

    • one year ago
  13. klimenkov Group Title
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    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?

    • one year ago
  14. mukushla Group Title
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    one question...i want make myself sure :) \[x_i>1\]?

    • one year ago
  15. klimenkov Group Title
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    I have a way to prove this using the inequality for arithmetic and geometric means.

    • one year ago
  16. Ted.knows.it Group Title
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    now i was looking at that idea. it seems the right one...

    • one year ago
  17. Ted.knows.it Group Title
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    @mukushla , yes it is given by the function domain

    • one year ago
  18. mukushla Group Title
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    thank u :)

    • one year ago
  19. klimenkov Group Title
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    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.

    • one year ago
  20. Ted.knows.it Group Title
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    yeah, so all in all, the problem goes down to proving that arithmetic mean is bigger than the geometric mean..

    • one year ago
  21. klimenkov Group Title
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    Yes. I wish you luck while proving this one.

    • one year ago
  22. Ted.knows.it Group Title
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    thanks for your time. :)

    • one year ago
  23. mukushla Group Title
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    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

    • one year ago
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