<s>[INST] Prove this Generalizing AM-GM inequality 
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:
$$a^3+b^3+c^3+3abc\ge ab(a+b)+bc(b+c)+ac(a+c)$$
 [/INST] Maybe this can help you, it seems very similar. 


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