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show that $$\sum\limits_{i=1}^n \frac{x_i}{i^2} \geq \frac{1}{1} + \frac{1}{2} + \dots +\frac{1}{n}$$ where $x_1,x_2,\dots,x_n$ are natural numbers and all of them are different numbers(no such a $x_i=x_j$) the teacher said you can prove it by making it a Cauchy form inequality.

thing i have tried to make Cauchy inequality and show it's same as question inequality:

multiply left side by $(1^2+2^2...+n^2)$.

multiply right side by $$\sum\limits_{i=1}^n \frac{i^2}{x_i}$$

and in none of them i was successful.

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  • $\begingroup$ The inequality is not true as stated. $\endgroup$ – N. S. Jul 11 '14 at 13:50
  • $\begingroup$ sorry. i fixed it. $\endgroup$ – user2838619 Jul 11 '14 at 13:54
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A proof just with Cauchy-Schwarz:

From Cauchy-Schwarz, one have that $$ (\sum_i \frac{1}{x_i})(\sum_i \frac{x_i}{i^2}) \ge (\sum_i \frac{1}{i})^2. $$ But since $\sum_i \frac{1}{x_i} \le \sum_i \frac{1}{i}$, one have that $$ \sum_i \frac{x_i}{i^2} \ge \frac{(\sum_i \frac{1}{i})^2}{\sum_i \frac{1}{x_i}} \ge \frac{\sum_i \frac{1}{i}}{\sum_i \frac{1}{x_i}} \sum_i \frac{1}{i} \ge \sum_i\frac{1}{i}. $$

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By the Rearrangement Inequality (but we don't need anything that general) the left side is minimized, for fixed $x_i$, if the $x_i$ are increasing. And then the minimum is reached if the $x_i$ are as small as possible, which gives $x_i=i$.

Remark: If we don't want to quote the Rearrangement Inequality, it is clear that if $i\lt j$ and $x_i \gt x_j$, then $\frac{x_i}{i^2}+\frac{x_j}{j^2} \gt \frac{x_j}{i^2}+\frac{x_i}{j^2}$.

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  • $\begingroup$ thanks for your nice and short solution. i never thought the question could be solved this easy.if there was no obligation on using Cauchy i would definitely accepted your answer. $\endgroup$ – user2838619 Jul 11 '14 at 14:24
  • $\begingroup$ You are welcome. The answer you accepted is precisely the answer to your problem as posed. Mine is not. $\endgroup$ – André Nicolas Jul 11 '14 at 14:27
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To give - in Addition to Andres answer - a solution using the Cauchy-Schwarz inqquality, we let $\xi_i := \frac{\sqrt{x_i}}i$, and $\eta_i := \frac{1}{\sqrt{x_i}}$. Then, by Cauchy-Schwarz \begin{align*} \sum_i \frac 1i &= \sum_{i} \xi_i \eta_i \\ &\le \left(\sum_{i} \xi_i^2\right)^{1/2} \left(\sum_i \eta_i^2\right)^{1/2}\\ &= \left(\sum_i \frac{x_i}{i^2}\right)^{1/2} \left(\sum_i \frac 1{x_i}\right)^{1/2} \end{align*} Now choose a $\sigma \in S_n$ such that $x_{\sigma(1)} < \ldots < x_{\sigma(n)}$, then $x_{\sigma(i)}\ge i$, giving $$ \sum_i \frac 1{x_i} = \sum_j \frac 1{x_{\sigma(j)} }\le \sum_j \frac 1j $$ Continuing above $$ \sum_i \frac 1i \le \left(\sum_i \frac{x_i}{i^2}\right)^{1/2} \left(\sum_i \frac 1i\right)^{1/2} $$ Hence diving by $\left(\sum \frac 1i\right)^{1/2}$ and squaring gives the result.

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  • $\begingroup$ thank you for your nice solution. i choose Mohammad answer because it was simpler to understand for me.i will up vote it when i got enough reputation. $\endgroup$ – user2838619 Jul 11 '14 at 14:19

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