<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] By Chebyshev's sum inequality (https://en.wikipedia.org/wiki/Chebyshev%27s_sum_inequality),
we have
$$a^n\cdot a + b^n\cdot b + c^n \cdot c
\ge \frac{1}{3}(a^n + b^n + c^n)(a+b+c).$$
Thus,
$$\frac{a^{n+1}+b^{n+1} + c^{n+1}}{a^n + b^n + c^n} \ge \frac{a+b+c}{3} \ge \sqrt[3]{abc}.$$
We are done.
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