Prove this Generalizing AM-GM inequality

Question

Let $n\ge 2$ and $a_{i} \ge 0,i=1,2,\cdots,n$, show that $$(n-1)^{n-1}(a^n_{1}+a^n_{2}+\cdots+a^n_{n})+n^na_{1}a_{2}\cdots a_{n}\ge (a_{1}+a_{2}+\cdots+a_{n})^n$$

When $n=2$, $$a^2_{2}+a^2_{2}+4a_{1}a_{2}=(a_{1}+a_{2})^2+2a_{1}a_{2}\ge (a_{1}+a_{2})^2$$

When $n=3$, it is $$4(a^3_{1}+a^3_{2}+a^3_{3})+27a_{1}a_{2}a_{3}\ge (a_{1}+a_{2}+a_{3})^3$$ By $$(a_{1}+a_{2}+a_{3})^3=a^3_{1}+a^3_{2}+a^3_{3}+3a_{1}a_{2}(a_{1}+a_{2})+3a_{1}a_{3}(a_{1}+a_{3})+3a_{2}a_{3}(a_{2}+a_{3})+6a_{1}a_{2}a_{3}$$

so it's enough to prove

$$a^3_{1}+a^3_{2}+a^3_{3}+7a_{1}a_{2}a_{3}\ge a_{1}a_{2}(a_{1}+a_{2})+a_{1}a_{3}(a_{1}+a_{3})+a_{2}a_{3}(a_{2}+a_{3})$$ which is clear by using Schur inequality: $$a^3+b^3+c^3+3abc\ge ab(a+b)+bc(b+c)+ac(a+c)$$

Answer 1

Maybe this can help you, it seems very similar.

Answer 2

We can use the Vasc's EV-method. See here the corollary 1.7 (b):

https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06_www.pdf

Indeed, let $a_1\leq a_2\leq...\leq a_n$, $a_1+a_2+...+a_n=const$ and $a_1^n+a_2^n+...+a_n^n=const$.

Thus, by EV $a_1a_2...a_n$ gets a minimal value in the following cases.

1. One of our variables is equal to $0$.

Let $a_n=0$.

In this case we need to prove that $$(n-1)^{n-1}\sum_{k=1}^{n-1}a_k^n\geq\left(\sum_{k=1}^{n-1}a_k\right)^n$$ or $$\frac{\sum\limits_{k=1}^{n-1}a_k^n}{n-1}\geq\left(\frac{\sum\limits_{k=1}^{n-1}a_k}{n-1}\right)^n,$$ which is Power Mean inequality;

2. $a_1=x$ and $a_2=...=a_n=1$, where $0\leq x\leq1$.

We need to prove that $f(x)\geq0$, where $$f(x)=(n-1)^{n-1}(x^n+n-1)+n^nx-(x+n-1)^n.$$ But $$f'(x)=n(n-1)^{n-1}x^{n-1}+n^n-n(x+n-1)^{n-1}\geq n^n-n\cdot (1+n-1)^{n-1}=0,$$ which says that $f$ is an increasing function.

Thus, $f(x)\geq f(0)=0$ and we are done!