If $x_i \gt 0$ for $1\leq i\leq n$ and $x_1+...+x_n=\pi$ Then the greatest value of $\sin x_1+...+\sin x_n$ is $n \sin \left(\frac{\pi}{n}\right)$. If $x_i \gt 0$ for $1\leq i\leq n$ and $x_1+x_2+x_3+...+x_n=\pi$
Then the greatest value of 
$\sin x_1+\sin x_2+...+\sin x_n$ is $n \sin \left(\frac{\pi}{n}\right)$.
Prove it.
I have no idea how to prove it. Please help.
 A: Hint: Use Jensen's Inequality as $\sin x$ is concave in the interval.
A: Hint: For $n \ge 1$, suppose $(x_1,\ldots,x_n)$ yielded the maximum value of $\sin x_1 + \cdots + \sin x_n$ subject to $x_1 + \cdots + x_n = \pi$, and that $x_i \neq x_j$ for some $i,j$. Then, we have: 
$\sin x_i + \sin x_j = 2\sin\left(\tfrac{x_i+x_j}{2}\right)\cos\left(\tfrac{x_i-x_j}{2}\right) < 2\sin\left(\tfrac{x_i+x_j}{2}\right) = \sin\left(\tfrac{x_i+x_j}{2}\right)+\sin\left(\tfrac{x_i+x_j}{2}\right)$. 
Can you get a contradiction from this? 
If so, then the maximum of $\sin x_1 + \cdots + \sin x_n$ subject to $x_1 + \cdots + x_n = \pi$ is attained when all the $x_i$'s are equal, i.e. $x_i = \frac{\pi}{n}$.
Of course, you need to show that $0 \le \cos\left(\tfrac{x_i-x_j}{2}\right) < 1$ first.
A: $\sin{x}$ is a strictly concave function on $[0,\pi]$, so $\sin{\frac{x+y}{2}}\gt\frac{\sin{x}+\sin{y}}{2}$ when $x\neq y$. 
If $\exists$ $i$ and $j$ s.t. $x_i\neq x_j$, then we can replace $x_i,x_j$ with $x^{'}_i,x^{'}_j$, where $x^{'}_i=x^{'}_j=\frac{x_i+x_j}{2}$, and we would have $\sin{x^{'}_i}+\sin{x^{'}_j}=2\sin{\frac{x_i+x_j}{2}}\gt\sin{x_i}+\sin{x_j}$. So, the maximum has to be reached when all $x_i$ are the same. 
Specifically, it is reached when $x_i=\frac{\pi}{n}$ for all $i$, and the maximum is $n\sin{\frac{\pi}{n}}$.
