# Making a proof using AM-GM.

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.

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 $$\frac{a_1}{b_1} \cdot \frac{a_2}{b_2} \cdots \frac{a_n}{b_n}$$?
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.