# Proving an inequality about a sequnce with Cauchy-Schwarz

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.

#### Solutions Collecting From Web of "Proving an inequality about a sequnce with Cauchy-Schwarz"

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

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.