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.

Just to make it clear, this is not really an answer, more of a failed approach using an obvious idea to try
We will assume $n \geq 3$, since otherwise the statement does not make a lot of sense (for $n = 2$ we would get $x_1 + x_2 = \pi$ so $\sin(x_1+x_2) = 0$ in the denominator, and similarly for $n = 1$).
First note that since $x_i \gt 0$ and $x_1 + x_2 + \cdots + x_n = \pi$, we must have $0 \lt x_i \lt \pi$ and $0 \lt x_{i+1} + x_i \lt \pi$ for all $i = 1, \ldots n$. Therefore we also have $\sin x_i \gt 0$ and $\sin(x_{i+1} + x_i) \gt 0$ so each term is strictly positive.
Taking logs we have
$$\log\left[\frac{\sin{x_{1}}\sin{x_{2}}\cdots\sin{x_{n}}}{\sin{(x_{1}+x_{2})}\sin{(x_{2}+x_{3})}\cdots\sin{(x_{n}+x_{1})}}\right]$$ $$ = \log\left[\frac{\sin x_1}{\sin(x_1+x_2)}\right]+\log\left[\frac{\sin x_2}{\sin(x_2+x_3)}\right]+\cdots+\log\left[\frac{\sin x_n}{\sin(x_n+x_1)}\right]$$
Now consider the function $f(x, y) = \log\left[\frac{\sin x}{\sin(x + y)}\right] = \log\left(\sin x\right) - \log\left[\sin\left(x+y\right)\right]$ on the domain satisfying the constraints $0 \lt x \lt \pi$, $0 \lt y \lt \pi$ and $0 \lt x + y \lt \pi$. We can compute the Hessian matrix and check whether it is negative-semidefinite on this domain, which is equivalent to $f$ being concave.
Assuming this was true, we could make use of Jensen's inequality for concave functions: $$\frac{f(x_1, x_2) + f(x_2, x_3) + \cdots + f(x_n, x_1)}{n} \leq f\left(\frac{x_1+x_2+\cdots+x_n}{n},\frac{x_1+x_2+\cdots+x_n}{n}\right)$$
This would imply $$\log\left[\frac{\sin{x_{1}}\sin{x_{2}}\cdots\sin{x_{n}}}{\sin{(x_{1}+x_{2})}\sin{(x_{2}+x_{3})}\cdots\sin{(x_{n}+x_{1})}}\right] \leq n \log\left[\frac{\sin\frac{\pi}{n}}{\sin\frac{2\pi}{n}}\right]$$ 
and the result would follow after taking exponents.
So what's left is to actually compute the Hessian matrix and check the criterion for negative-semidefiniteness, and also make sure that Jensen's inequality holds in this way for multivariable functions.
Now we find $\frac{\partial f}{\partial x} = \cot x - \cot(x+y)$ and $\frac{\partial f}{\partial y} = -\cot(x+y)$. Then we compute $$\frac{\partial^2 f}{\partial x^2} = \csc^2(x+y)-\csc^2 x$$ $$\frac{\partial^2 f}{\partial x\partial y} = \frac{\partial^2 f}{\partial y\partial x} = \csc^2(x+y)$$ $$\frac{\partial^2 f}{\partial y^2} = \csc^2(x+y)$$
Hence the Hessian matrix is given by $$\begin{pmatrix}
\csc^2(x+y) - \csc^2 x & \csc^2(x+y) \\
\csc^2(x+y) & \csc^2(x+y)
\end{pmatrix}$$
Its determinant is $-\csc^2(x+y)\csc^2 x$ so it is strictly negative in the domain we're working in. This means that the eigenvalues have opposite signs, and so the function is neither concave nor convex, which means that Jensen's inequality should not hold, and there ought to be counterexamples to the claim.
One issue is that we are restricted in the kinds of points we can look at (they must all share a common component with another point, like $(x_1, x_2), (x_2, x_3), \cdots, (x_n, x_1)$) so it's conceivable that there might not be counterexamples that satisfy this additional constraint. I am not really sure how to deal with this at the moment.
