Making a proof using AM-GM.

Question

Theorem: Let $a_1,a_2,a_3,\cdots a_n$ be a sequence of positive numbers and let $b_1,b_2,\cdots, b_n$ be any permutation of the first sequence. Then $$\frac{a_1}{b_1}+\frac{a_2}{b_2}+\cdots +\frac{a_n}{b_n} \ge n$$

I'm not sure how to proceed or even the meaning of the question. If anyone could provide some insight it would be highly appreciated. I'm a first-year math student who's in a little over her head.

Answer 1

Hint: Use AM-GM on the sequence $\frac{a_1}{b_1},\frac{a_2}{b_2},\dots,\frac{a_n}{b_n}$. Because $b_1,b_2,\dots,b_n$ is any permutation of $a_1,a_2,\dots,a_n$, what can you say about the product $\frac{a_1}{b_1} \cdot \frac{a_2}{b_2} \cdots \frac{a_n}{b_n}$?

Solution: (hover over the yellow box to see it)

First, note that because $b_1,b_2,\dots,b_n$ is any permumatation of $a_1,a_2,\dots,a_n$ we must have that $$\frac{a_1a_2\cdots a_n}{b_1b_2\cdots b_n} = 1$$ By AM-GM we have $$\frac{\frac{a_1}{b_1} + \frac{a_2}{b_2} + \dots + \frac{a_n}{b_n}}{n} \geq \sqrt[n]{\frac{a_1}{b_1} \cdot \frac{a_2}{b_2} \cdots \frac{a_n}{b_n}} = \sqrt[n]{1} = 1$$ Now multiply by $n$ on both sides, and the inequality you wish to prove follows.