How to prove $\frac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt[3]{abc}$? Give $a,b,c>0$. Prove that: $$\dfrac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt[3]{abc}.$$
My direction: (we have the equation if and only if $a=b=c$)
$a^{n+1}+a^nb+a^nc \ge 3a^n\sqrt[3]{abc}$
$b^{n+1}+b^na+b^nc \ge 3b^n\sqrt[3]{abc}$
$c^{n+1}+c^na+c^nb \ge 3c^n\sqrt[3]{abc}$
But from these things, i can't prove the problem.

Let $A_p:=\left(\frac{1}{N}\sum_{i=1}^Na_i^p\right)^{1/p}$ be the $p$th mean of $(a_i)$.
By the extended AM-GM inequality, $GM\le A_n\le A_{n+1}$. Hence $$GM\times A_n^n\le A_{n+1}\times A_{n+1}^n=A_{n+1}^{n+1}$$ or $$\sqrt[3]{abc}\times\frac{a^n+b^n+c^n}{3}\le\frac{a^{n+1}+b^{n+1}+c^{n+1}}{3}$$
