Find the maximum of the value$(n^{n-1}-1)\sqrt[n]{a_{1}a_{2}\cdots a_{n}}+\sqrt[n]{\frac{a^n_{1}+a^n_{2}+\cdots+a^n_{n}}{n}}$ 
Let $n$ be give a positive integer, $a_{i}
\ge 0$,such that $a_{1}+a_{2}+\cdots+a_{n}=n$. Find the maximum value of
  $$(n^{n-1}-1)\sqrt[n]{a_{1}a_{2}\cdots a_{n}}+\sqrt[n]{\dfrac{a^n_{1}+a^n_{2}+\cdots+a^n_{n}}{n}}$$

Using AM-GM we have
$$\sqrt[n]{a_{1}a_{2}\cdots a_{n}}\le\dfrac{a_{1}+a_{2}+\cdots+a_{n}}{n}=1$$
But other hand we have
$$a^n_{1}+a^n_{2}+\cdots+a^n_{n}\ge\dfrac{1}{n^{n-1}}(a_{1}+a_{2}+\cdots+a_{n})^n=\dfrac{1}{n^{n-1}}$$
 A: dezdichado's answer is correct, I'll just write out all the computations.
Clearly the maximum value is $n^{n-1}$.
Let $X = \frac{1}{n}\sum a_i^n$ and $Y=\prod a_i$. We would like to show that $\left(n^{n-1}-1\right)\sqrt[n]{Y}+\sqrt[n]{X}\le n^{n-1}$
Expanding and using AM-GM: $$n^n=\left(\sum a_i\right)^n=\sum a_i^n+\sum_{|\alpha|=n} \left(\begin{matrix}n\\ \alpha\end{matrix}\right)a^\alpha\ge nX+\left(n^n-n\right)Y$$ So $$n^{n-1}\ge X+\left(n^{n-1}-1\right)Y $$
Now applying Hölder's inequality gives us: $$\left[X+\left(n^{n-1}-1\right)Y\right]\left(1+n^{n-1}-1\right)^{n-1}\ge \left(\sqrt[n]{X}+\left(n^{n-1}-1\right)\sqrt[n]{Y}\right)^n$$
Simplifying that results in $$n^{n-1}n^{(n-1)^2}=\left(n^{n-1}\right)^n\ge \left(\sqrt[n]{X}+\left(n^{n-1}-1\right)\sqrt[n]{Y}\right)^n$$
Which completes the proof. So the maximum value is $n^{n-1}$ when all $a_i$ are equal to $1$
A: Hint: Expand $n^n = (a_1+a_2+...+a_n)^n = a_1^n+a_2^n+...+a_n^n+\sum a_1^{k_1}a_2^{k_2}...a_n^{k_n}$ and now use AM-GM on the last $n^n-n$ terms. 
