<s>[INST] 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.
 [/INST] Because for natural $n$ by AM-GM we obtain: $$\begin{aligned} \sum_{\text{cyc}}a^{n+1}&=\frac{1}{3(n+1)}\sum_{\text{cyc}}\left((3n+1)a^{n+1}+b^{n+1}+c^{n+1}\right) \\ 
&\geq \sum_{\text{cyc}}\sqrt[3n+3]{a^{(3n+1)(n+1)}b^{n+1}c^{b+1}} \\
&=\sum_{\text{cyc}}a^{n+\frac{1}{3}}b^{\frac{1}{3}}c^{\frac{1}{3}} \\
&=\sqrt[3]{abc}\sum_{\text{cyc}}a^n \end{aligned}$$
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