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$$ \prod\limits_{k=1}^{n}a_{k} < {1 \over n^{n}}\left(\,\sum_{k = 1}^{n}\,\sqrt{1+a_{k}\,a_{k+1}\,}\,\right)^n $$

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.

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  • 1
    $\begingroup$ What are $a_k$ and $x_k$? $\endgroup$ – Zubin Mukerjee Dec 3 '13 at 16:31
  • $\begingroup$ I correct it, x is just a, problem solved. $\endgroup$ – haia Dec 3 '13 at 16:56
  • $\begingroup$ But what is $a_k$? Just any real number? (not all $a_i$ will make the square root on the right well defined) $\endgroup$ – Zubin Mukerjee Dec 3 '13 at 17:06
  • $\begingroup$ To attack the problem, the first is to figure out what are the allowed ranges of $a_k$. Since there is a square root, it might make sense to assume all $a_k > 0$. However, $a_k$ cannot be arbitrary big. This is because if you scale everything by a big $\lambda$, the LHS growth as $\lambda^n$ while the RHS growth as $\lambda^2$ only. This means the inequality will fail for large enough $a_k$. Back to square one, what is your $a_k$? $\endgroup$ – achille hui Dec 3 '13 at 17:25
  • $\begingroup$ ak is a real number $\endgroup$ – haia Dec 3 '13 at 17:29
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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

$$ \begin{align} \prod_{k=1}^n a_k &= \prod_{k=1}^n \sqrt{a_k a_k} = \sqrt{a_1}\left(\prod_{k=1}^{n-1}\sqrt{a_k a_{k+1}}\right)\sqrt{a_n}\\ &= \prod_{k=1}^n \sqrt{a_k a_{k+1}} \quad\color{blue}{\longleftarrow a_{1} = a_{n+1} \text{ and rearrange }}\\ &\le\left(\frac{\sum_{k=1}^n \sqrt{a_k a_{k+1}}}{n}\right)^n \quad\color{blue}{\longleftarrow \text{GM} \le \text{AM}}\\ &< \left(\frac{\sum_{k=1}^n \sqrt{1+a_k a_{k+1}}}{n}\right)^n = \frac{1}{n^n}\left(\sum_{k=1}^n \sqrt{1+a_k a_{k+1}}\right)^n \end{align} $$

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  • $\begingroup$ It's very nice! +1 $\endgroup$ – math110 Dec 5 '13 at 2:35

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