shobber wrote: Let $g(x)= \sum_{k=1}^{n} a_k \cos{kx}$, $a_1,a_2, \cdots, a_n, x \in R$. If $g(x) \geq -1$ holds for every $x \in R$, prove that $\sum_{k=1}^{n}a_k \leq n$. The hint see here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=65041

please post a perfect solution of this problem.thanks

shobber wrote: Let $g(x)= \sum_{k=1}^{n} a_k \cos{kx}$, $a_1,a_2, \cdots, a_n, x \in R$. If $g(x) \geq -1$ holds for every $x \in R$, prove that $\sum_{k=1}^{n}a_k \leq n$. Putting $x_j:=\frac{j\pi}{n+1}$, $j=1,...,n$ and adding it all up yields \[-n\le\sum_{j=1}^ng(x_j)=\sum_{j=1}^n\sum_{k=1}^na_k \cos{\frac{jk\pi}{n+1}}=-\sum_{k=1}^na_k\] using $\sum_{j=0}^n \cos{\frac{jk\pi}{n+1}=0}$. (see $e^{2k\pi i}-1=0$)