Show $\frac{\sin x_1\sin x_2\cdots\sin x_n}{\sin(x_1+x_2)\sin(x_2+x_3)\cdots\sin(x_n+x_1)}\le\frac{\sin^n(\pi/n)}{\sin^n(2\pi/n)}$, for $\sum x_i=\pi$ Let $x_{i}>0$, ($i=1,2,\cdots,n$) and such that
$$x_{1}+x_{2}+\cdots+x_{n}=\pi.$$
Show that
$$
\dfrac{\sin{x_{1}}\sin{x_{2}}\cdots\sin{x_{n}}}{\sin{(x_{1}+x_{2})}\sin{(x_{2}+x_{3})}\cdots\sin{(x_{n}+x_{1})}}\le\left(\dfrac{\sin{\frac{\pi}{n}}}{\sin{\frac{2\pi}{n}}}\right)^n
$$
This problem also post  MO,Until now No one solve it,I think there might be a solution here, because I 've heard that there are a lot of people here who are good at and like inequality, so the possibility of solving this inequality is very high, and I really look forward to them.

Update:
I have been thinking about this more.
The idea is to replace $x_i$ with some M such that all the properties holds and the ratio of the $\sin$ product holds and the ratio of the products of sines is smaller than the ratio formed with $\sin(M)$.  Initially I thought that selecting the Maximum among $x_i$ is sufficient but since $x_{max} + x_{max} > \pi$ as per comments from @Sangchul. This is insufficient. 
Consider this,
We have $\sin(x_i + x_{i+1}) \gt 0$, $\sin(x_i) \gt 0$ and $0 \lt x_i + x_{i+1} < \pi ---->(0)$
$$\left(\frac{x_1 + x_2 + x_3+ ...+x_n}{n}\right)^n \ge x_1x_2...x_n ----> (1)$$ 
Also, we have for $k > 0$
$$\left(x_1 + x_2 + x_3 + ...+ x_n\right)^k \ge x_1^k + x_2^k + x_3^k + ..+x_n^k --> (2)$$
for all positive reals.
This can be used. 
$x_1 + x_2 + x_3 + ...x_n = \pi$
Now from ($0$),($1$) and ($2$) $$\frac{\sin{x_1}\sin{x_2}\sin{x_3}...\sin{x_n}}{\sin({x_1 + x_2})\sin({x_2 + x_3})\sin({x_3 + x_4})...\sin({x_{n} + x_{1}})}$$
$$\le \frac{\left(\frac{\sin{x_1} + \sin{x_2}...+\sin{x_n}}{n}\right)^n}{\sin({x_1 + x_2})\sin({x_2 + x_3})\sin({x_3 + x_4})...\sin({x_{n} + x_{1}})}$$
$$\le \frac{\left(\sin{x_1} + \sin{x_2}...+\sin{x_n}\right)^n}{n^n\sin({x_1 + x_2})\sin({x_2 + x_3})\sin({x_3 + x_4})...\sin({x_{n} + x_{1}})}$$ 
Now, use MacLaurin's  series expansion of $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!}-\frac{x^7}{7!}...$
$$\le \dfrac{\left(\left((x_1 + x_2 + x_3 + ...+x_n) + \frac{(x_1^5 + x_2^5 + x_3^5+ ...+x_n^5)}{5!} + \frac{(x_1^9 + x_2^9 + x_3^9+ ...+x_n^9)}{9!} + ...\right) - \left(\frac{(x_1^3 + x_2^3 + x_3^3+ ...+x_n^3)}{3!} + \frac{(x_1^7 + x_2^7 + x_3^7+ ...+x_n^7)}{7!} + ... \right)\right)^n}{n^n\sin({x_1 + x_2})\sin({x_2 + x_3})\sin({x_3 + x_4})...\sin({x_{n} + x_{1}})}$$ 
Now, using ($2$) and keeping the denominator same, we have a ratio that is bigger.
$$\le \dfrac{\left(\left((x_1 + x_2 + x_3 + ...+x_n) + \frac{(x_1 + x_2 + x_3 + ...+x_n)^5}{5!} + \frac{(x_1 + x_2 + x_3+ ...+x_n)^9}{9!} + ...\right) - \left(\frac{(x_1 + x_2 + x_3+ ...+x_n)^3}{3!} + \frac{(x_1 + x_2 + x_3+ ...+x_n)^7}{7!} + ... \right)\right)^n}{n^n\sin({x_1 + x_2})\sin({x_2 + x_3})\sin({x_3 + x_4})...\sin({x_{n} + x_{1}})}$$.
But then,  $x_1 + x_2 + x_3 + ...x_n = \pi$, so we have
$$\le \dfrac{\left(\left(\pi + \frac{\pi^5}{5!} + \frac{\pi^9}{9!} + ...\right) - \left(\frac{(\pi^3}{3!} + \frac{\pi^7}{7!} + ... \right)\right)^n}{n^n\sin({x_1 + x_2})\sin({x_2 + x_3})\sin({x_3 + x_4})...\sin({x_{n} + x_{1}})}$$
$$\le \dfrac{(\sin \pi)^n}{<denominator>}$$
$$\le 0 !!$$
This means the assumption about the positivity of 
$$\frac{\sin{x_1}\sin{x_2}\sin{x_3}...\sin{x_n}}{\sin({x_1 + x_2})\sin({x_2 + x_3})\sin({x_3 + x_4})...\sin({x_{n} + x_{1}})}$$
Seems Incorrect?
