Find the limit $ \lim_{n \to \infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n$ For $a,b,c>0$, Find $$ \lim_{n \to \infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n$$
how can I find the limit of sequence above? 
Provide me a hint or full solution.
thanks ^^

This answer is inspired by @Aryabhata's answer here. 
Suppose $x_k>0$. 
By $\mathrm{AM}\ge\mathrm{GM}$ we have 
$$\sqrt[m]{x_1\cdots x_m} \le \frac{x_1+\ldots +x_m}{m}.$$
By weighted $\mathrm{AM}\ge\mathrm{GM}$ we have 
$$\begin{eqnarray*}
\frac{m}{x_1+\ldots +x_m} 
&=& \frac{x_1\cdot 1/x_1+\ldots +x_m\cdot 1/x_m}{x_1+\ldots +x_m} \\
%&\ge& \left(\frac{1}{x_1^{x_1}\cdots x_m^{x_m}}\right)^{1/(x_1+\ldots +x_m)} \\
%&\ge& \left(\frac{1}{x_1^{x_1}\cdots x_m^{x_m}}\right)^{\frac{1}{x_1+\ldots +x_m}} \\
&\ge& \sqrt[x_1+\ldots +x_m]{\frac{1}{x_1^{x_1}\cdots x_m^{x_m}}}.
\end{eqnarray*}$$
Therefore, 
$$\sqrt[m]{x_1\cdots x_m} 
\le \frac{x_1+\ldots +x_m}{m}
\le \sqrt[x_1+\ldots +x_m]{x_1^{x_1}\cdots x_m^{x_m}}.$$
Letting $x_k = a_k^{1/n}$ and taking the $n$th power we find by the squeeze theorem that 
$$\lim_{n \to \infty}\left(\frac{a_1^{1/n}+\ldots +a_m^{1/n}}{m}\right)^n
= \sqrt[m]{a_1\cdots a_m}.$$
In particular, 
$$\lim_{n \to \infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n
= \sqrt[3]{a b c}.$$ 
