It may be a strengthening form of mean inequality

Let $a_{i}>0,i=1,2,\cdots,n>2$,show that following inequality $$\prod_{i=1}^{n}\left(\dfrac{\displaystyle\sum_{j\neq i}a_{j}}{n-1}\right)^{n-1}\ge\left(\dfrac{\displaystyle\sum_{i=1}^{n}\prod_{j\neq i}a_{j}}{n}\right)^n$$

I have found that this inequality feeling can be handled using mean inequality, but I have, finally, found it difficult to handle. Thanks.

For $n=3$ use well konwn $9-8$inequality $$9(a+b)(b+c)(c+a)\ge 8(a+b+c)(ab+bc+ca)$$ so we have $$(a+b)^2(b+c)^2(a+c)^2\ge \dfrac{64}{81}(a+b+c)^2(ab+bc+ac)^2$$ and $$(a+b+c)^2\ge 3(ab+bc+ca)$$ so we have $$(a+b)^2(b+c)^2(a+c)^2\ge \dfrac{64}{27}(ab+bc+ca)^3$$ so we have $$\left(\dfrac{(a+b)}{2}\right)^2\left(\dfrac{b+c}{2}\right)^2\left(\dfrac{c+a}{2}\right)^2\ge\left(\dfrac{ab+bc+ca}{3}\right)^3$$

Solution 1:

Let $\sum\limits_{i=1}^na_i=nu_1,$ $\sum\limits_{1\leq i<j\leq n}a_ia_j=\binom{n}{2}u_2^2,$ ..., $\prod\limits_{i=1}^na_i=u_n^n,$ where $u_k>0$ for any $k$.

Thus, by the Maclaurin's inequality we obtain: $$u_1\geq u_2\geq...\geq u_n$$ and we need to prove that $$\prod_{i=1}^n\left(\frac{nu_1-a_i}{n-1}\right)^{n-1}\geq u_{n-1}^{n(n-1)}$$ or $$\prod_{i=1}^n(nu_1-a_i)\geq(n-1)^nu_{n-1}^n$$ or $$n^nu_1^n-n^{n-1}u_1^{n-1}\cdot nu_1+n^{n-2}\binom{n}{2}u_1^{n-2}u_2^2+...+(-1)^n\prod_{i=1}^na_i\geq(n-1)^nu_{n-1}^n$$ or $$\sum_{i=2}^n(-1)^in^{n-i}\binom{n}{i}u_1^{n-i}u_i^i\geq(n-1)^nu_{n-1}^n.$$ Now, consider two cases.

$n$ is odd.

Since the equality in our inequality occurs for $a_1=a_2=...=a_n=1,$ we obtain: $$\sum_{i=2}^n(-1)^in^{n-i}\binom{n}{i}=(n-1)^n.$$ Now for any natural $k$ such that $2\leq k\leq n$ we see that $$\sum_{i=2}^k(-1)^in^{n-i}\binom{n}{i}=\left(n^{n-2}\binom{n}{2}-n^{n-3}\binom{n}{n-3}\right)+...+$$ $$+\left(n^{n-2j}\binom{n}{2j}-n^{n-2j-1}\binom{n}{2j+1}\right)+...>0$$ because $$n^{n-2j}\binom{n}{2j}-n^{n-2j-1}\binom{n}{2j+1}>0$$ it's $$\frac{n\cdot n!}{(2j)!(n-2j)!}>\frac{n!}{(2j+1)!(n-2j-1)!}$$ or $$n(2j+1)>n-2j$$ or $$2j(n+1)>0,$$ which is obvious.

Thus, by Maclaurin we obtain: $$\sum_{i=2}^n(-1)^in^{n-i}\binom{n}{i}u_1^{n-i}u_i^i\geq\left((n-1)^n+1\right)u_{n-1}^{n-1}u_1-u_n^n\geq(n-1)^nu_{n-1}^n;$$

$n$ is even.

Since for all even $n>4$ we have $$7<n^{n-2}\binom{n}{2}-n^{n-3}\binom{n}{3}<(n-1)^n-8,$$ by Maclaurin again we obtain: $$\sum_{i=2}^n(-1)^in^{n-i}\binom{n}{i}u_1^{n-i}u_i^i\geq7u_1^{n-2}u_2^2+\left((n-1)^n-8\right)u_1^2u_{n-2}^{n-2}+u_n^n.$$

Thus, it's enough to prove that $$7u_1^{n-2}u_2^2+\left((n-1)^n-8\right)u_1^2u_{n-2}^{n-2}+u_n^n\geq(n-1)^nu_{n-1}^n$$ for which it's enough to prove that: $$7u_1^{n-2}u_2^2+u_n^n\geq8u_{n-1}^n.$$ Now, for any even $n>4$ we have the following reasoning.

$a_i$ are positive roots of the equation: $$\prod_{i=1}^n(x-a_i)=0$$ or $$x^n-\binom{n}{1}u_1x^{n-1}+\binom{n}{2}u_2^2x^{n-2}+...+\binom{n}{n-2}u_{n-2}^{n-2}x^2-\binom{n}{n-1}u_{n-1}^{n-1}x+u_n^n=0,$$ which says that the equation $$u_n^nx^n-\binom{n}{1}u_{n-1}^{n-1}x^{n-1}+\binom{n}{2}u_{n-2}^{n-2}x^{n-2}-\binom{n}{3}u_{n-3}^{n-3}x^{n-3}+\binom{n}{4}u_{n-4}^{n-4}x^{n-4}-...+1=0$$ has $n$ positive roots and by the Roole's theorem we see that the equation $$\left(u_n^nx^n-\binom{n}{1}u_{n-1}^{n-1}x^{n-1}+\binom{n}{2}u_{n-2}^{n-2}x^{n-2}-\binom{n}{3}u_{n-3}^{n-3}x^{n-3}+\binom{n}{4}u_{n-4}^{n-4}x^{n-4}-...+1\right)^{(n-4)}=0$$ or $$u_n^nx^4-4u_{n-1}^{n-1}x^3+6u_{n-2}^{n-2}x^2-4u_{n-3}^{n-3}x+u_{n-4}^{n-4}=0$$ has four positive roots, which says that the equation $$u_n^n-4u_{n-1}^{n-1}x+6u_{n-2}^{n-2}x^2-4u_{n-3}^{n-3}x^3+u_{n-4}^{n-4}x^4=0$$ has four positive roots.

Let $a$, $b$, $c$ and $d$ be roots of the last equation and let $$a+b+c+d=4u,$$ $$ab+ac+bc+ad+bd+cd=6v^2,$$ $$abc+abd+acd+bcd=4w^3$$ and $$abcd=t^4.$$ Thus, $$u=\frac{u_{n-3}^{n-3}}{u_{n-4}^{n-4}},$$ $$v^2=\frac{u_{n-2}^{n-2}}{u_{n-4}^{n-4}},$$ $$w^3=\frac{u_{n-1}^{n-1}}{u_{n-4}^{n-4}}$$ and $$t^4=\frac{u_{n}^{n}}{u_{n-4}^{n-4}}.$$ Now, $$(a-b)^2(c-d)^2+(a-c)^2(b-d)^2+(a-d)^2(b-c)^2\geq0$$ it's $$t^4\geq4uw^3-3v^4,$$ which gives $$\frac{u_{n}^{n}}{u_{n-4}^{n-4}}\geq4\frac{u_{n-3}^{n-3}}{u_{n-4}^{n-4}}\cdot\frac{u_{n-1}^{n-1}}{u_{n-4}^{n-4}}-3\left(\frac{u_{n-2}^{n-2}}{u_{n-4}^{n-4}}\right)^2.$$
Id est, it's enough to prove that: $$7u_1^{n-2}u_2^2+\frac{4u_{n-3}^{n-3}u_{n-1}^{n-1}}{u_{n-4}^{n-4}}-\frac{3u_{n-2}^{2n-4}}{u_{n-4}^{n-4}}\geq8u_{n-1}^n,$$ which is true because by AM-GM and Maclaurin $$4u_1^{n-2}u_2^2+\frac{4u_{n-3}^{n-3}u_{n-1}^{n-1}}{u_{n-4}^{n-4}}\geq8\sqrt{\frac{u_1^{n-2}u_2^2u_{n-3}^{n-3}u_{n-1}^{n-1}}{u_{n-4}^{n-4}}}\geq8\sqrt{\frac{u_{n-4}^{n-4}u_{n-1}^2u_{n-1}^2u_{n-1}^{n-3}u_{n-1}^{n-1}}{u_{n-4}^{n-4}}}=8u_{n-1}^n$$ and $$3u_1^{n-2}u_2^2-\frac{3u_{n-2}^{2n-4}}{u_{n-4}^{n-4}}\geq0$$ it's $$u_1^{n-2}u_2^2u_{n-4}^{n-4}\geq u_{n-2}^{2n-4},$$ which is obvious by Maclaurin.