TITLE: Proving an inequality about a sequnce with Cauchy-Schwarz
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: 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
$$
\sum_i \frac{x_i}{i^2} \ge 
\frac{(\sum_i \frac{1}{i})^2}{\sum_i \frac{1}{x_i}}  \ge 
\frac{\sum_i \frac{1}{i}}{\sum_i \frac{1}{x_i}} \sum_i \frac{1}{i}  \ge 
\sum_i\frac{1}{i}.
$$