Sum on the ratio of elements in a set and its permuation

Question

lxhfcjksdnbcjk s dkjbdjksehdkin swdwskdnbwsdkjwsnbdkwnsknw

Answer 1

A direct application of the AM-GM inequality gives:

$\sum_{i=1}^n\frac{a_i}{b_i}\geq n\sqrt[n]{\prod_{i=1}^n\frac{a_i}{b_i}}=n$,

since $\{b\}_{i=1}^n$ is a permutation of $\{a\}_{i=1}^n$, all the terms in the product cancel out.

Answer 2

Chebyshev inequality says that if we have two sets of $n$ positive numbers $(x_k)$ in increasing order and $(y_k)$ (not necessarily ordered) and a permutation $\sigma \in S_n$ then the sum $\sum_{k=1}^n x_ky_{\sigma(k)}$ is minimal when the sequece $y_{\sigma(k)}$ is decreasing. In other terms, the sum of combined products is minimal when the orderings are opposite.

Now in this problem we can assume that the $(a_i)$s are ordered increasingly. By Chebyshev's theorem the sum $\sum a_i\frac{1}{b_i}$ is minimal when the $b_i$s are ordered such that $1/b_i$ forms a decreasing sequence. This is only possible when $b_i = a_i$ and thus the minimal sum is equal to $n$.