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 
$$a^2+b^2+c^2+3(abc)^{2/3}\ge 2(ab+bc+ca)$$
 A: 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.
A: A dull proof for $n<5$ by BW(Buffalo Way).
For $n=4$, it is enough to prove$$64(a^3b+b^3c+c^3d+d^3a)+(a+b+c+d)^2 \ge 32(a+b+c+d)(abc+abd+acd+bcd)$$and WLOG we can assume $a$ is smallest of the four numbers.
That is, $b=a+u$, $c=a+v$ and $d=a+w$ for some positive numbers $u,v,w$. Expanding gives
$$64a^2(3 u^2 -  u v - 4 u w + 3v^2 - v w + 3 w^2)\\
+16a\left(\left(\sum_{cyc} 5u^3-u^2w\right) + 11 u^2 v - 14 u v w -  u w^2 + 11 v^2 w  \right)\\
+\left(\left(\sum_{cyc} u^4+6u^2v^2+4u^3w-20u^2vw\right) + 68 u^3 v + 4 u w^3 + 68 v^3 w \right)\ge 0$$
and each coeffcients are nonnegative. Indeed,
$$3 u^2 -  u v - 4 u w + 3v^2 - v w + 3 w^2 \ge 2(u-w)^2+0.5(u-v)^2+0.5(v-w)^2 \ge 0,$$
$$\left(\sum_{cyc} 5u^3-u^2w\right) + 11 u^2 v - 14 u v w -  u w^2 + 11 v^2 w\\\ge \left(\sum_{cyc} 4u^3\right) + 4 u^2 v  - u w^2 -7vw^2 + 11 v^2 w \\\ge (4u^3-uw^2+2w^3)+(4v^3+11v^2w-7vw^2+2w^3)\ge 0$$
and
$$\left(\sum_{cyc} u^4+6u^2v^2+4u^3w-20u^2vw\right) + 68 u^3 v + 4 u w^3 + 68 v^3 w\\\ge\left(\sum_{cyc} 4u^3w-13u^2vw\right) + 68 u^3 v + 4 u w^3 + 68 v^3w\\\ge v(68u^3-13u^2w-13uw^2+4w^3)+w(68v^3-13uv^2+4u^3)\ge 0$$
For $n=3$, it is enough to show
$$27(a^3b+b^3c+c^3a)+(a+b+c)^4\ge6(ab+bc+ca)(a+b+c)^2$$
and similarly, let $b=a+u$, $c=a+v$ and expanding gives
$$45a^2(u^2 -  u v + v^2)\\
+9a(3 u^3 + 5 u^2 v - 4 u v^2 + 3 v^3)\\
+(u^4 + 25 u^3 v - 6 u^2 v^2 - 2 u v^3 + v^4)\ge 0$$
and you can check each of these polynomials are nonnegative for positive $u, v, w$.
