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.

Observe that by AM-GM inequality for $3$ positive reals: $\dfrac{a+b+c}{3} \ge \sqrt[3]{abc}$. Thus you need to show: $\dfrac{a^{n+1}+b^{n+1} + c^{n+1}}{a^n+b^n+c^n} \ge \dfrac{a+b+c}{3}$, which is the same as: $2(a^{n+1}+b^{n+1}+c^{n+1}) \ge a^n(b+c)+b^n(c+a)+c^n(a+b)$. But this is quite clear because you can use the inequality:
$(x-y)(x^n-y^n) \ge 0 \implies x^{n+1}+y^{n+1} \ge x^ny+xy^n$ three times for the pairs $(a,b), (b,c), (c,a)$ and add up.
