$$ \prod\limits_{k=1}^{n}a_{k} < {1 \over n^{n}}\left(\,\sum_{k = 1}^{n}\,\sqrt{1+a_{k}\,a_{k+1}\,}\,\right)^n $$
ak and n are positive real number greater than 0.
EDIT: a_{k+1} becomes a_{1} when a_{k}=a_{n}, it is a cylic notation. SORRY.
Any ideas of how to attack the problem?? Thank You.
I don't know if this could help, but the 1/n is also the exponent for the left hand side. I'm thinking maybe of log??
I'm pretty sure that at some point it would be helpful the binominal coefficent?? I don't know.
After the question get cleared up, the answer becomes a trivial application of GM $\le$ AM. Since the $a_k$ are in cyclic notation. i.e. $a_{n+1} = a_1$, we have
$$ \begin{align} \prod_{k=1}^n a_k &= \prod_{k=1}^n \sqrt{a_k a_k} = \sqrt{a_1}\left(\prod_{k=1}^{n-1}\sqrt{a_k a_{k+1}}\right)\sqrt{a_n}\\ &= \prod_{k=1}^n \sqrt{a_k a_{k+1}} \quad\color{blue}{\longleftarrow a_{1} = a_{n+1} \text{ and rearrange }}\\ &\le\left(\frac{\sum_{k=1}^n \sqrt{a_k a_{k+1}}}{n}\right)^n \quad\color{blue}{\longleftarrow \text{GM} \le \text{AM}}\\ &< \left(\frac{\sum_{k=1}^n \sqrt{1+a_k a_{k+1}}}{n}\right)^n = \frac{1}{n^n}\left(\sum_{k=1}^n \sqrt{1+a_k a_{k+1}}\right)^n \end{align} $$