<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] WLOG let $a \geq b \geq c$ , then $\sqrt[3]a \geq \sqrt[3]b \geq \sqrt[3]c$. Let $x = \sqrt[3]a, y = \sqrt[3]b , z =\sqrt[3]c$ , then $x \geq y\geq z > 0$ and the inequality is equivalent to :
$$
x^{3n+3} + y^{3n+3} + z^{3n+3} \geq x^{3n+1}yz + y^{3n+1}xz + z^{3n+1}xy   
$$
by Muirhead's inequality, since the sequence $(3n+3,0,0)$ majorizes $(3n+1,1,1)$, we know that
$$
\sum_{\mathrm{sym}} x^{3n+3}y^0z^0 \geq \sum_{\mathrm{sym}} x^{3n+1}y^1z^1
$$
which upon division by $2$ gives the result we require. Note that $n$ natural is required for the inequality to hold.
</s>