Find $\sigma$ for which $\sum _{i=1}^{n} \frac{1}{a_i\cdot a_{\sigma(i)}}$ is maximal

Question

$$\text{Let: } 0 < a_1 \lt a_2 \lt \dots \lt a_n$$

$$\text{Find } \sigma \in S_n \text{ for which :}$$

$$\sum _{i=1}^{n} \frac{1}{a_i\cdot a_{\sigma(i)}} \text{ is maximal}$$

I think the maximum value of the sum will be reached if $\sigma = e$ but i don't know how to prove that. Any idea is really appreciated! Thanks for help!

Note: $e = \begin{pmatrix} 1 & 2 & ... & n\\ 1 & 2 & ... & n \end{pmatrix}$

Answer 1

By Cauchy Schwarz inequality $$\sum_{i=1}^n \frac{1}{a_ia_{\sigma(i)}} \leq \left(\sum_{i=1}^n\frac{1}{a_i^2}\right)^{1/2}\left(\sum_{i=1}^n\frac{1}{a_{\sigma(i)}^2}\right)^{1/2}=\sum_{i=1}^n\frac{1}{a_i^2}$$ The equality occurs when $a_i=\lambda a_{\sigma(i)}$ for all $i$. This would mean that $$\prod_ia_i=\lambda^n \prod_{i}a_{\sigma(i)}=\lambda^n\prod_ia_i \implies \lambda=1$$

Thus $\sigma$ is the identity permutation.