Prove that: $ \sum_{1}^{n}{ \cos^2{x_i}} \le n\cos^2{\frac{\pi}{2n}}$ I am looking for a proof of the problem following:
Let $0 < x_i < \pi/2$ for $i=1, 2, 3,..., n$ and $\sum_{1}^{n}{ x_i=\pi/2}$ . 
Prove that:  $ \sum_{1}^{n}{ \cos^2{x_i}} \le n\cos^2{\frac{\pi}{2n}}$
 A: First let's note that the restriction $0 < x_k < \pi/2$ combined with $\sum_{k=1}^n x_k = \pi/2$ means that we must have $n \ge 2$.
Equality is clearly obtained when $x_1 = x_2 = \cdots = x_n = \frac{\pi}{2n}$.  I think, and maybe I'm wrong, that it can be shown using multivariable calculus that this equality is actually the maximum.  At the very least we can show it's a maximum.  I don't know if multivariable calc is above your level or not and I generally don't like to provide answers above the OP's level, but I'm not sure how else to get started (maybe induction?) and it's an interesting problem.
Here's an outline of how to get started using multivariable calculus:


*

*Define $f(x_1, x_2, \dots, x_n) = \sum_{k=1}^n \cos^2(x_k)$.

*Note that $f_{x_k} = -\sin(2x_k)$ for all $k$.

*Therefore $f_{x_{kk}} = - 2\cos(2x_k)$ for all $k$ and $f_{x_{kj}} = 0$ for all $j \ne k$.

*This means the Hessian matrix has a structure so simple that finding its eigenvalues is trivial.

*If $n \ge 3$ and if $x_1 = x_2 = \cdots = x_n = \frac{\pi}{2n}$ then the Hessian matrix is positive definite.  This means that $f$ has a local max at $(x_1, \dots, x_n) = (\frac{\pi}{2n}, \dots, \frac{\pi}{2n})$.


It still remains to show that this is actually the global max.  The case $n = 2$ also need to be considered.  Note that $n=2$ won't work in step 5 above because then the Hessian matrix is the zero matrix and we can't draw any conclusions from that analysis.
A: I think we can use Jensen's inequality. $\cos^{2}$ is concave on $0 < x < \frac{\pi}{2}$ and therefore 
  $$ \cos^{2} \left( \frac{1}{n} \sum x_{i} \right) \geq \frac{1}{n} \sum \cos^{2}(x_{i})$$
