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.

  • $\begingroup$ 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

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 $$ \sum_i \frac{x_i}{i^2} \ge \frac{(\sum_i \frac{1}{i})^2}{\sum_i \frac{1}{x_i}} \ge \frac{\sum_i \frac{1}{i}}{\sum_i \frac{1}{x_i}} \sum_i \frac{1}{i} \ge \sum_i\frac{1}{i}. $$


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 $\frac{x_i}{i^2}+\frac{x_j}{j^2} \gt \frac{x_j}{i^2}+\frac{x_i}{j^2}$.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.