<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] Another way.
We need to prove that:
$$\sum_{cyc}a^{n+1}\geq\sum_{cyc}a^{n+\frac{1}{3}}b^{\frac{1}{3}}c^{\frac{1}{3}},$$ which is true by Muirhead because $$(n+1,0,0)\succ\left(n+\frac{1}{3},\frac{1}{3},\frac{1}{3}\right).$$
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