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Prove that if $x_1, ..., x_{n-1}$ are positive numbers and $n \geq 2$, than the following inequality holds:

$x_1^2 + x_2^3 + ... + x_{n - 1}^n + \frac{1}{x_1^2 x_2^3 ... x_{n - 1}^n} \geq n + (x_1 - 1)^2 + 2(x_2 - 1)^2 + ... + (n - 1)(x_{n - 1} - 1)^2$.

Firstly, I tried to prove this inequality by induction, but it didn't work. Then I tried using the AM-GM inequality, and finally some substitutions, but none of them lead me to a nicer form.

Do you have any suggestions for this inequality?

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  • $\begingroup$ And what about $n$? Is $$n\geq 1$$? $\endgroup$ – Dr. Sonnhard Graubner Dec 31 '18 at 10:51
  • $\begingroup$ Could you show your work, having proceeded as you mention? $\endgroup$ – Namaste Dec 31 '18 at 21:52
  • $\begingroup$ This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc. $\endgroup$ – Carl Mummert Jan 11 at 13:55
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For all $n\geq2$ we need to prove that $$\sum_{k=1}^{n-1}(x_k^{k+1}-k(x_k-1)^2)+\frac{1}{\prod\limits_{k=1}^{n-1}x_k^{k+1}}\geq n$$ or $$\sum_{k=1}^{n-1}(x_k^{k+1}-kx_k^2+2kx_k)+\frac{1}{\prod\limits_{k=1}^{n-1}x_k^{k+1}}\geq\frac{n(n-1)}{2}+n$$ or $$\sum_{k=1}^{n-1}(x_k^{k+1}-kx_k^2+(k-1)x_k)+\sum_{k=1}^{n-1}(k+1)x_k+\frac{1}{\prod\limits_{k=1}^{n-1}x_k^{k+1}}\geq\frac{n(n+1)}{2},$$ which is true by AM-GM: $$x_k^{k+1}-kx_k^2+(k-1)x_k=k_k(x_k^k+(k-1)\cdot1-kx_k)\geq0$$ and $$\sum_{k=1}^{n-1}(k+1)x_k+\frac{1}{\prod\limits_{k=1}^{n-1}x_k^{k+1}}\geq\frac{n(n+1)}{2}\left(\prod\limits_{k=1}^{n-1}x_k^{k+1}\cdot\frac{1}{\prod\limits_{k=1}^{n-1}x_k^{k+1}}\right)^{\frac{2}{n(n+1)}}=\frac{n(n+1)}{2}.$$

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  • $\begingroup$ Thank you a lot, this was extremely nice and helpful! And I wish you a happy New Year! :) $\endgroup$ – Sandel Dec 31 '18 at 17:49
  • $\begingroup$ Happy New Year! $\endgroup$ – Michael Rozenberg Dec 31 '18 at 17:50

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