Prove that for any rearrangement of $b_1, b_2, \cdots, b_n$ of the positive numbers $a_1,a_2,\cdots,a_n$ one has $\sum\limits_{i=1}^n \dfrac{a_i }{b_i} \geq n$.
Attempt:
First, one can write $\sum \frac{a_i}{b_i} = \sum \dfrac{a_i + b_i }{b_i} - n \geq n$ so we can prove
$$ \sum \dfrac{ a_i + b_i }{b_i} \geq 2n \iff \sum \dfrac{a_i + b_i }{2b_i} \geq n $$
and we a sum of $n$ numbers of the form $\dfrac{ a_i + b_i }{2b_i} \geq \dfrac{\sqrt{a_i b_i } }{ b_i} = \sqrt{ \dfrac{ a_i}{b_i} }$. I see a no-end from here since it is not always true that $a_i \geq b_i$.
....
Other strategy is perhaps use induction? If $n=2$, then we have $(a_1,a_2)$ an if $b_i = a_i$ then we have the result easily. If $b_1 = a_2$, then
$$ \dfrac{a_1}{a_2} + \dfrac{a_2}{a_1} \geq 2 $$
which follows because $x + 1/x \geq 2 $. Now, suppose result is true for $n$ then
$$ \sum^{n} \dfrac{a_i}{b_i} + \dfrac{a_{n+1}}{b_{n+1}} \geq n + \dfrac{a_{n+1}}{b_{n+1} } $$
how can one prove that $a_{n+1} \geq b_{n+1}$ ?
