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Let $n \in \mathbb{N}$. How to find the least positive integer $\gamma$ such that $ \sum\limits_{i=1}^{n} \cos{\theta_i} \leq \gamma$ provided $\prod\limits_{i=1}^{n} \tan{\theta_i} = 2^{\frac{n}{2}}$ for any $\theta_i \in (0,\frac{\pi}{2})$, $i=1, 2, \ldots, n$?

I had trouble solving this problem, some time ago. If I am not mistaken, I had seen this in Loney's trigonometry book. (Although I don't remember it correctly!).

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Try this. Find the maximum of $\sum_{k=1}^n \frac{1}{\sqrt{1+{x_k}^2}},$ where $\prod_{k=1}^n {x_k}^2=2^n,$ and $x_k = \tan\theta_k$, using Lagrange multipliers. Each $x_k$ satisfies the same equation involving the multiplier $\lambda,$ $\lambda 2^{n+1} = \frac{{x_k}^2}{(1+{x_k}^2)^{3/2}}.$ And so all the $x_k$ are equal and hence $x_k = \sqrt{2}.$ Therefore the maximum value of our sum is $n/\sqrt{3}$ and so the required smallest integer is $\lceil \frac{n}{\sqrt{3}} \rceil.$

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    When $n=2$, I found that $\gamma=2$. The maximal value for $\cos\theta_1+\cos\theta_2$ is around 1.1547. However, for n>2, I don't seem to find examples for now with a value over $n-1$.2011-12-25