Let us consider $a_1,a_2,\cdots a_n>0$ such that $a_1 a_2\cdots a_n=1$ then show that $a_1^{n-1}+a_2^{n-1}+\cdots a_n^{n-1}\geq \frac{1}{a_1}+\frac{1}{a_2}+\cdots \frac{1}{a_n}$.
Attempt 1(definition)
Since the the inequality is equal to $(\frac{a_1^n}{a_1}-\frac{1}{a_1})+(\frac{a_2^n}{a_2}-\frac{1}{a_2})+\cdots +(\frac{a_n^n}{a_n}-\frac{1}{a_n})\geq 0$ but the inequality $$\frac{1}{a_1}(a_1^n-1)\geq 0$$ can´t be true for each $a_i$.
Attempt 2 (by substitution)
Now from our hypothesis $\frac{1}{a_i}=a_1a_2\cdots \hat{a_i} a_{i+1}\cdots a_n$ and putting it in our inequality $a_1^n(a_2a_3\cdots a_n)+a_2^n(a_1a_3\cdots a_n)+\cdots + a_n^n(a_1a_2\cdots a_{n-1})\geq (a_2a_3\cdots a_n)+(a_1a_3\cdots a_n)+\cdots +(a_1a_2\cdots a_{n-1})$
but each $a_i>0$ and hence $\frac{1}{a_i}>0$ and then our assertion is true.
Attempt 3 (rearrangement inequality)
WLOG assume that $a_1\leq a_2\cdots \leq a_n$ I need find a collection of ordered numbers $b_1\leq b_2\cdots b_n$ for use that $a_1b_1+a_2b_2+\cdots a_nb_n\geq a_1b_{\sigma(1)}+a_2b_{\sigma(2)}+\cdots+a_nb_{\sigma(n)}$ where $\sigma\in S_n$.
But I cant figure out how should be $b_i$ for each $i$.
Any advice or hint was helpful.
