<s>[INST] show this inequality with $\sum_{i=1}^{n}a_{i}=n$ 
Let $n\ge 3$ be postive  number,$a_{i}>0,i=1,2,\cdots,n$,and $\displaystyle\sum_{i=1}^{n}a_{i}=n$,show that
  $$a^3_{1}a_{2}+a^3_{2}a_{3}+\cdots+a^3_{n}a_{1}+n\ge 2(a_{1}a_{2}\cdots a_{n-1}+a_{2}a_{3}\cdots a_{n}+a_{n}a_{1}\cdots a_{n-2})$$

it seem can use indution to prove it.when $n=3$,it must prove
$$a^3_{1}a_{2}+a^3_{2}a_{3}+a^3_{3}a_{1}+3\ge 2(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1})$$
it seem  use three shcur inequaliy 
$$a^3+b^3+c^3+3abc\ge \sum ab(a+b)$$
then we have 
$$a^2+b^2+c^2+3(abc)^{2/3}\ge 2(ab+bc+ca)$$
 [/INST] The hint.
Prove this inequality for $n=3$ and for $n=4$.
But for all $n\geq5$ by AM-GM we obtain:
$$\sum_{k=1}^na_k^3a_{k+1}\geq n\sqrt[n]{\prod_{k=1}^na_k^4}.$$
Thus, it's enough to prove that:
$$n\sqrt[n]{\prod_{k=1}^na_k^4}+n\geq2\prod_{k=1}^na_k\sum_{k=1}^n\frac{1}{a_k}.$$
Now, let $\prod\limits_{k=1}^na_k=const$ and $f(x)=-\frac{1}{x}.$
Thus, $$g(x)=f'\left(\frac{1}{x}\right)=x^2$$ is strictly convex on $(0,+\infty),$ which says that by Vasc's EV Method, theorem 1.3 (1), $p=0$ from here
https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf
the expression $\sum\limits_{k=1}^nf(x_k)=-\sum\limits_{k=1}^n\frac{1}{a_k}$ gets a minimal value, when $n-1$ variables are equal.
After homogenization it's enough to assume that $a_1=a$ and $a_2=...=a_n=1$ and we need to prove that:
$$n\sqrt[n]{a^4}\left(\frac{a+n-1}{n}\right)^{n-5}+n\left(\frac{a+n-1}{n}\right)^{n-1}\geq2((n-1)a+1),$$
which is true by AM-GM!
Indeed, let $a=x^n$.
Thus, we need to prove that
$$P(x)=x^{n^2-n}+b_1x^{n^2-n-1}+...+b_{n^2-n}\geq0,$$ where only coefficient before $x^n$ is negative.
But the inequality $$n\sqrt[n]{\prod_{k=1}^na_k^4}+n\geq2\prod_{k=1}^na_k\sum_{k=1}^n\frac{1}{a_k}.$$
 is symmetric, which says that the polynomial $P$ is divisible by $(x-1)^2.$
Now, we see that the sum of coefficients of $P'(x)$ is equal to zero, 
which says that $P(x)\geq0$ by AM-GM.
</s>