# Prove that for positive $a_n$, and $b_n$ any rearrangement of $a_n$, one has $\sum \frac{a_i}{b_i} \geq n$

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 $$a_{n+1} \geq b_{n+1}$$ ?

• Maybe just a direct result of the rearrangement inequality?
– Yuta
Commented May 8, 2020 at 3:34
• Just apply the AM-GM inequality to the fractions $\frac{a_k}{b_k}$. Commented May 8, 2020 at 3:40
• Got it! because $\sum \dfrac{a_i}{b_i} \geq n ( \prod \frac{a_i}{b+i} )^{1/n} = n$. Correct? Thanks profesor! Write it as an answer to reward you for your reply! Commented May 8, 2020 at 5:06