# Show that $\frac{a_1}{b_1}+\frac{a_2}{b_2}+...+\frac{a_n}{b_n} \geq n$

Let $a_1$, $a_2$,..., $a_n$ be the sequence of positive numbers, and let $b_1$, $b_2$,..., $b_n$ be any permutation of the first sequence. Show that $$\frac{a_1}{b_1}+\frac{a_2}{b_2}+...+\frac{a_n}{b_n} \geq n$$

This is a preparation Question PUTNAM contest. The theme of the section is "inequality". I can not make the problem from inequality. Is anyone is able give me a hint (using inequalities)?

## 2 Answers

Hint: AM-GM. Note that the product of our fractions is $1$.

It's quite simple, but just to clarify even more:

$$\dfrac{\dfrac{a_1}{b_1} + \dfrac{a_2}{b_2} + ... + \dfrac{a_n}{b_n}}{n} \geq \sqrt[n]{\dfrac{a_1}{b_1}\dfrac{a_2}{b_2} ... \dfrac{a_n}{b_n}}$$

But $$\dfrac{a_1}{b_1}\dfrac{a_2}{b_2} ... \dfrac{a_n}{b_n} = 1$$

Then,

$$\dfrac{\dfrac{a_1}{b_1} + \dfrac{a_2}{b_2} + ... + \dfrac{a_n}{b_n}}{n}\geq 1$$

Then it follows that,

$$\dfrac{a_1}{b_1} + \dfrac{a_2}{b_2} + ... + \dfrac{a_n}{b_n}\geq n$$

I hoped to have made it better understood, and just to let the Geometric-Arithmetic inequality for general use, this is, Geometric Mean is always smaller equal than arithmetic mean.