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 $x_{i}&gt;0$, ($i=1,2,\cdots,n$) 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 1

1) I found that for $n=4, 5$, although there are infinitely many stationary points, 
each stationary point is the global maximizer. In other words, the objective function is constant for all stationary points. Is this true for $n &gt; 5$?

Take $n=4$ for example.
Let
$$g(x_1, x_2, x_3) = \frac{\sin x_1\sin x_2 \sin x_3 \sin (x_1+x_2+x_3)}{(\sin (x_1+x_2))^2 (\sin (x_2+x_3))^2} ,$$
\begin{align}
f(x_1, x_2, x_3) = \ln g(x_1, x_2, x_3)  &amp;= \ln \sin x_1 + \ln \sin x_2 + \ln \sin x_3 + \ln \sin (x_1 + x_2 + x_3)\nonumber\\
&amp;\qquad - 2\ln \sin (x_1+x_2) - 2\ln \sin (x_2 + x_3).
\end{align}
The stationary points are those feasible points with $\frac{\partial f}{\partial x_1} = \frac{\partial f}{\partial x_2} = \frac{\partial f}{\partial x_3} = 0.$
I found that $$\frac{\partial f}{\partial x_1} = \frac{\partial f}{\partial x_2} = \frac{\partial f}{\partial x_3} = 0 \Longrightarrow
g(x_1, x_2, x_3) = \frac{1}{4}.$$

In detail, we have
\begin{align}
\cot x_1 + \cot (x_1 + x_2 + x_3) - 2\cot (x_1 + x_2) &amp;= 0, \qquad (1)\\
\cot x_2 + \cot (x_1 + x_2 + x_3) - 2\cot (x_1 + x_2) - 2\cot (x_2 + x_3) &amp;= 0,\\
\cot x_3 + \cot (x_1 + x_2 + x_3) - 2\cot (x_2 + x_3) &amp;= 0.
\end{align}
By letting $u_1 =\cot x_1, \ u_2 = \cot x_2, \ u_3 = \cot x_3$,
we have $(1) \Longrightarrow u_1u_2 + u_2u_3 + u_3u_1 - u_2^2 - 2 = 0$.
On the other hand, $g(x_1, x_2, x_3) - \frac{1}{4} = \frac{(u_1u_2+u_2u_3+u_3u_1-u_2^2 - 2)^2}{4(u_1+u_2)^2(u_2+u_3)^2} = 0.$

2) We can see this from another view.

Case $n=4$: Let $u_i = \cot x_i, \ i=1,2,3$. We have (noting that $x_4 = \pi - x_1-x_2-x_3$)
\begin{align}
&amp;\frac{\sin x_1 \sin x_2 \sin x_3 \sin x_4}{\sin(x_1+x_2)\sin (x_2+x_3)\sin (x_3+x_4)\sin (x_4+x_1)}\nonumber\\
=\ &amp; \frac{\sin x_1\sin x_2 \sin x_3 \sin (x_1+x_2+x_3)}{(\sin (x_1+x_2))^2 (\sin (x_2+x_3))^2}\nonumber\\
=\ &amp; \frac{(u_1u_2 + u_2u_3+u_3u_1-1)(1+u_2^2)}{(u_1+u_2)^2(u_2+u_3)^2}\nonumber\\
=\ &amp; \frac{1}{4} - \frac{(u_1u_2+u_2u_3+u_3u_1-u_2^2 - 2)^2}{4(u_1+u_2)^2(u_2+u_3)^2}.
\end{align}

Case $n=5$: Let $u_i = \cot x_i, \ i=1,2,3, 4$. We have (noting that $x_5 = \pi - x_1-x_2-x_3-x_4$)
\begin{align}
&amp;\frac{\sin x_1 \sin x_2 \sin x_3 \sin x_4 \sin x_5}{\sin(x_1+x_2)\sin (x_2+x_3)\sin (x_3+x_4)\sin (x_4+x_5)\sin (x_5+x_1)}\nonumber\\
=\ &amp; \frac{\sin x_1 \sin x_2 \sin x_3 \sin x_4 \sin (x_1+x_2+x_3+x_4)}{\sin(x_1+x_2)\sin (x_2+x_3)\sin (x_3+x_4)\sin (x_1+x_2+x_3)\sin (x_2+x_3+x_4)}\nonumber\\
=\ &amp; \frac{(u_1u_2u_3 + u_1u_2u_4 + u_1u_3u_4 + u_2u_3u_4 - u_1-u_2-u_3-u_4)(1+u_2^2)(1+u_3^2)}
{(u_1+u_2)(u_2+u_3)(u_3+u_4)(u_1u_2+u_2u_3+u_3u_1-1)(u_2u_3+u_3u_4+u_4u_2-1)}.\quad (2)
\end{align}
Denote (2) as $f(u_1, u_2, u_3, u_4)$.
It follows from $\frac{\partial f}{\partial u_4} = 0$
that $u_1 = g(u_2, u_3, u_4)$. Let $h(u_2, u_3, u_4) = f(g(u_2, u_3, u_4), u_2, u_3, u_4)$.
It follows from $\frac{\partial h}{\partial u_4} = 0$ that $u_2 = F(u_3, u_4)$.
Then $h(F(u_3, u_4), u_3, u_4) = \frac{5\sqrt{5}-11}{2}.$
Remark: Here $g, h, F$ are some rational functions whose expressions are not given, for the sake of simplicity.

Update

Proof of $n=4$:

We need to prove that
$$\frac{\sin x_1 \sin x_2 \sin x_3 \sin x_4}{\sin(x_1+x_2)\sin (x_2+x_3)\sin (x_3+x_4)\sin (x_4+x_1)} \le \Big(\frac{\sin \frac{\pi}{4}}{\sin\frac{\pi}{2}}\Big)^4.$$
It suffices to prove that
$$(\sin(x_1+x_2))^2(\sin(x_2+x_3))^2 - 4\sin x_1 \sin x_2 \sin x_3 \sin(x_1+x_2+x_3) \ge 0.$$
Using substitutions $$\cos x_1 = \frac{1-w_1^2}{1+w_1^2}, \ \sin x_1 = \frac{2w_1}{1+w_1^2}, \ \cos x_2 = \frac{1-w_2^2}{1+w_2^2}, \ \sin x_2 = \frac{2w_2}{1+w_2^2}, \\
\cos x_3 = \frac{1-w_3^2}{1+w_3^2}, \ \sin x_3 = \frac{2w_3}{1+w_3^2},$$ the inequality becomes
$$\frac{16 Q^2}{(w_1^2+1)^2 (w_2^2+1)^4 (w_3^2+1)^2}\ge 0$$
where 
\begin{align}
Q &amp;= w_1^2 w_2^3 w_3+w_1^2 w_2^2 w_3^2-w_1 w_2^4 w_3+w_1 w_2^3 w_3^2-w_1^2 w_2^2-w_1^2 w_2 w_3-w_1 w_2^3-6 w_1 w_2^2 w_3\nonumber\\
&amp;\qquad -w_1 w_2 w_3^2-w_2^3 w_3-w_2^2 w_3^2+w_1 w_2-w_1 w_3+w_2^2+w_2 w_3.
\end{align}
We are done.
Remark: We can prove $n=4$ without using above substitutions. For $n=5$, it is not so simple.

Previously written

This is not an answer. I want to point out that for $n=4, 5$, there exist infinitely many feasible points such that equality occurs. In other words, if the inequality holds, there exist infinitely many global maximizers.

1) $n=4$.

Let $x_1, x_2 \in (0, \frac{\pi}{2})$ satisfying
\begin{align}
(\cot x_1)^2 + 2\cot x_1 \cot x_2 - (\cot x_2)^2 -2 = 0.
\end{align}
Remark: We may solve $x_1$ from (1), that is, $x_1 = \mathrm{arccot}\frac{\sqrt{2}-\cos x_2}{\sin x_2}, \ x_2 \in (0, \frac{\pi}{2}).$

Let $x_3 = x_1,\ x_4 = \pi - x_1 - x_2 - x_3.$ We have $x_1, x_2, x_3, x_4 &gt; 0; \ x_1 + x_2 + x_3 + x_4 = \pi$ and
\begin{align}
&amp;\frac{\sin x_1 \sin x_2 \sin x_3 \sin x_4}{\sin(x_1+x_2)\sin (x_2+x_3)\sin (x_3+x_4)\sin (x_4+x_1)} - \Big(\frac{\sin \frac{\pi}{4}}{\sin\frac{\pi}{2}}\Big)^4\\
=\ &amp; \frac{(\sin x_1)^2\sin x_2 \sin (2x_1 + x_2)}{(\sin (x_1+x_2))^4} - \frac{1}{4}\\
=\ &amp; - \frac{((\cot x_1)^2 + 2\cot x_1 \cot x_2 - (\cot x_2)^2 -2)^2}{4(\cot x_1 + \cot x_2)^4}\\
=\ &amp;0.
\end{align}

2) $n = 5$.

Let $x_1, x_2 \in (0, \frac{\pi}{2})$ satisfying
\begin{align}
-4(\cot x_2)^2(\cot x_1)^2 + (-2(\cot x_2)^3 + 6\cot x_2)\cot x_1 + (\cot x_2)^4  + 4(\cot x_2)^2 - 1 = 0.
\end{align}
Let $y_1 = \cot x_1, \ y_2 = \cot x_2$. We have $-4y_2^2y_1^2 + (-2y_2^3 + 6y_2) y_1
+y_2^4+4y_2^2-1 = 0$ which results in $\sqrt{5}y_2^2-4y_1y_2-y_2^2+\sqrt{5}+3 = 0$
since $x_1, x_2 \in (0, \frac{\pi}{2}).$

Let $x_3 = x_2, \ x_4 = x_1, \ x_5 = \pi - x_1 - x_2 - x_3 - x_4.$
We have $x_1, x_2, x_3, x_4, x_5 &gt; 0; \ x_1 + x_2+x_3+x_4+x_5=\pi.$
Note that $\big(\frac{\sin \frac{\pi}{5}}{\sin \frac{2\pi}{5}}\big)^5 = \frac{5\sqrt{5}-11}{2}$. We have
\begin{align}
&amp;\frac{\sin x_1 \sin x_2 \sin x_3 \sin x_4 \sin x_5}
{\sin (x_1+x_2) \sin (x_2 + x_3) \sin (x_3 + x_4) \sin (x_4+x_5) \sin (x_5+x_1)}
- \Big(\frac{\sin \frac{\pi}{5}}{\sin \frac{2\pi}{5}}\Big)^5\\
=\ &amp;\frac{(\sin x_1)^2(\sin x_2)^2\sin (2x_1 + 2x_2)}{(\sin (x_1+x_2))^2\sin 2x_2 (\sin (x_1 + 2x_2))^2} - \frac{5\sqrt{5}-11}{2}\\
=\ &amp;\frac{(y_1y_2-1)(y_2^2+1)^2}{y_2(y_1+y_2)(2y_1y_2+y_2^2-1)^2} - \frac{5\sqrt{5}-11}{2}\\
=\ &amp;-\frac{5\sqrt{5}-11}{16}\frac{
(\sqrt{5}y_2^2+2y_1y_2+3y_2^2+\sqrt{5}+1)(\sqrt{5}y_2^2-4y_1y_2-y_2^2+\sqrt{5}+3)^2}
{y_2(y_1+y_2)(2y_1y_2+y_2^2-1)^2}\\
=\ &amp;0.
\end{align}
This is just too long for a comment .

I can prove a particular case of the general case of a variant using Jensen's inequality .


  
    Let $x_i&gt;0$ with $1\leq i\leq n$ such that $\sum_{i=1}^{n}x_i=\pi$ with $|x_i-x_{i+1}|=\epsilon$  and $n=2q+1$ where $q$ is a natural number then we have:
    $$
\dfrac{\sin{x_{1}}\sin{x_{2}}\cdots\sin{x_{n}}}{\sin{(x_{1}+x_{2})}\sin{(x_{2}+x_{3})}\cdots\sin{(\frac{2\pi}{n}+n\epsilon)}}\le\left(\dfrac{\sin{(\frac{\pi}{n})}}{\sin{(\frac{2\pi}{n}+\epsilon)}}\right)^n
$$
    Where $\frac{\pi}{n}-q\epsilon&gt;0$
  


Proof we can rewrite the inequality as :

$$
\dfrac{\sin{(\frac{\pi}{n}-q\epsilon)}\sin{(\frac{\pi}{n}-(q-1)\epsilon)}\cdots\sin{(\frac{\pi}{n})}\cdots\sin{(\frac{\pi}{n}+q\epsilon})}{\sin{((\frac{2\pi}{n}-(2q-1)\epsilon))}\sin{((\frac{2\pi}{n}-(2q-3)\epsilon))}\cdots \sin(\frac{2\pi}{n}+n\epsilon)}\le\left(\dfrac{\sin{(\frac{\pi}{n})}}{\sin{(\frac{2\pi}{n}+\epsilon)}}\right)^n
$$

We make the function $$f(x)=\frac{\sin{(\frac{\pi}{n}-q\epsilon+x)}}{\sin{(\frac{2\pi}{n}-(2q-1)\epsilon+2x)}}$$

The function $$g(x)=\ln(f(x))$$ is concave on the  interval $[0,2q\epsilon]$ provided that $\sin{(\frac{\pi}{n}-q\epsilon+x)}\leq \sin{(\frac{2\pi}{n}-(2q-1)\epsilon+2x)}$  remains to apply the Jensen's inequality to the function we get :

$$g(0)+g(\epsilon)+g(2\epsilon)+\cdots+g(2q\epsilon)\leq \ln\left(\left(\dfrac{\sin{(\frac{\pi}{n})}}{\sin{(\frac{2\pi}{n}+\epsilon)}}\right)^n\right)$$

So we have showed :

$$
\dfrac{\sin{x_{1}}\sin{x_{2}}\cdots\sin{x_{n}}}{\sin{(x_{1}+x_{2})}\sin{(x_{2}+x_{3})}\cdots\sin{(x_{n}+x_{1})}}\le \frac{\sin{(\frac{2\pi}{n}+n\epsilon)}}{\sin{(\frac{2\pi}{n})}}
\left(\dfrac{\sin{(\frac{\pi}{n})}}{\sin{(\frac{2\pi}{n}+\epsilon)}}\right)^n $$

Furthermore I have checked numerically that we have :

$\exists \,\epsilon&gt;0$ such that :

$$ \frac{\sin{(\frac{2\pi}{n}+\epsilon)}}{\sin{(\frac{2\pi}{n})}}
\left(\dfrac{\sin{(\frac{\pi}{n})}}{\sin{(\frac{2\pi}{n}+\epsilon)}}\right)^n \le \left(\dfrac{\sin{(\frac{\pi}{n})}}{\sin{(\frac{2\pi}{n})}}\right)^n$$

If we swap the variables we get $n!$ functions of the kind $f(x)$ wich is a difficulty .  

Another way

I recall the Levinson's inequality (Theorem B):


  
    Let $f:I=[a,b]\to \mathbf{R}$ be a $3$-convex function , and $a_n$ , $b_n$ $\in [a,b]$ for $n=1,2,\cdots,n$ such that :
    $$\operatorname{max}(a_1\cdots a_n)\leq \operatorname{min}(b_1\cdots b_n), a_1+b_1=a_2+b_2=\cdots=a_n+b_n$$
    Then :$$\frac{\sum_{i=1}^{n}f(b_i)}{n}-f\Big(\frac{\sum_{i=1}^{n}b_i}{n}\Big)\geq \frac{\sum_{i=1}^{n}f(a_i)}{n}-f\Big(\frac{\sum_{i=1}^{n}a_i}{n}\Big)$$
  


A part of the initial inequality is an application of the Levinson Theorem since we have :

$f(x)=\ln(\sin(x))$  and $f'''(x)\geq 0$ on $]0,\frac{\pi}{2}]$

And $0&lt;b_i=x_i+x_{i+1}\leq \frac{\pi}{2}$ and $0&lt;a_i=x_i&lt;\frac{\pi}{2}$

With the condition :$$\operatorname{max}(a_1\cdots a_n)\leq \operatorname{min}(b_1\cdots b_n), a_1+b_1=a_2+b_2=\cdots=a_n+b_n$$