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 $\sum x_i=\pi$

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 $n > 5$?

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) &= \ln \sin x_1 + \ln \sin x_2 + \ln \sin x_3 + \ln \sin (x_1 + x_2 + x_3)\nonumber\\ &\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) &= 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) &= 0,\\ \cot x_3 + \cot (x_1 + x_2 + x_3) - 2\cot (x_2 + x_3) &= 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} &\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\\ =\ & \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\\ =\ & \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\\ =\ & \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} &\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\\ =\ & \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\\ =\ & \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 &= 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\\ &\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 $x_1, x_2, x_3, x_4 > 0; \ x_1 + x_2 + x_3 + x_4 = \pi$ and \begin{align} &\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\\ =\ & \frac{(\sin x_1)^2\sin x_2 \sin (2x_1 + x_2)}{(\sin (x_1+x_2))^4} - \frac{1}{4}\\ =\ & - \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}\\ =\ &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 $x_1, x_2, x_3, x_4, x_5 > 0; \ x_1 + x_2+x_3+x_4+x_5=\pi.$ 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} &\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\\ =\ &\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}\\ =\ &\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}\\ =\ &-\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}\\ =\ &0. \end{align}

This was a promising attempt using Lagrange multipliers and Implicit Function Theorem that proved to be insufficient.

Take the natural log from both sides and write the equivalent inequality: $$f (x_1,\dots x_n) = \sum\limits_{i=1}^n(\ln\sin x_i - \ln(\sin (x_i+x_{i+1}))\leq n\ln\dfrac{\sin\frac \pi n}{\sin\frac{2\pi}{n}}.$$ Using Lagrange multiplier, we want to maximize $f$ subject to : $$g(x_1,\dots x_n) = x_1+x_2+\dots x_n -\pi = 0.$$ Therefore, we want to solve for the system of $n+1$ equations: $$\nabla f = \lambda\nabla g$$ or in terms of the index $i:$ $$\cot x_i - \cot(x_{i-1}+x_i) - \cot(x_i+x_{i+1}) = \lambda.$$ We want to conclude that the only solution to these system of equations in the interval $(0,\pi)$ subject to $g = 0$ happens when $x_1 = x_2 =...=x_n = \dfrac{\pi}{n}$ and $\lambda$ accordingly. Since it is readily verified that above is actually a solution, we can make use of the Implicit Function Theorem to assert the uniqueness.

That is, let $x^0 = (\frac{\pi}{n}, \frac{\pi}{n},...\frac{\pi}{n})$ and $\lambda^0 = \cot\frac{\pi}{n} - 2\cot\frac{2\pi}{n}$ and define the equations: $$f_i(x_1,\dots x_n,\lambda):= \cot x_i - \cot(x_{i-1}+x_i) - \cot(x_i+x_{i+1}) - \lambda=0.$$ Notice that $f_i(x^0,\lambda^0) = 0$ and we are done if we verify that the Jacobian matrix: $$J(x_1,\dots x_n) = \begin{pmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2}& \dots \frac{\partial f_1}{\partial x_n}\\ \vdots &\ddots &\vdots \\ \frac{\partial f_n}{\partial x_1} & \frac{\partial f_n}{\partial x_2}& \dots \frac{\partial f_n}{\partial x_n} \end{pmatrix}$$ is non-singular at the point $(x^0, \lambda ^0).$ Luckily, our matrix is a rather simple circulant matrix for which determinant formulae are readily available: $$J(x^0)=\begin{pmatrix} a & b& 0&0 \dots 0&b\\ b&a&b&0\dots 0&0\\ 0&b&a&b\dots 0&0\\ \vdots &\vdots &\vdots&\ddots \vdots&\vdots \\ b&0&0&0\dots a&b \end{pmatrix},$$ where \begin{cases} a = \dfrac{\partial f_i}{\partial x_i}(x^0,\lambda^0) = \dfrac{2}{\sin^2\frac{2\pi}{n}} - \dfrac{1}{\sin^2\frac \pi n} = -\dfrac{2\cos(\frac{2\pi}{n})}{\sin^2(\frac{2\pi}{n})}\\ b = \dfrac{\partial f_i}{\partial x_{i-1}}(x^0,\lambda^0) = \dfrac{\partial f_i}{\partial x_{i+1}}(x^0,\lambda^0) = \dfrac{1}{\sin^2\frac{2\pi}{n}}. \end{cases} With the standard circulant matrix notation, we have $c_ 0 = a, c_{n-1} = c_1 = b$ and all other $c_j$ are zeros. As such, the eigenvalues are given as: $$\mu_j = c_0+c_{n-1}w_j+c_{n-2}w_j^2+\dots+c_1w_j^{n-1} = a+b(w_j+w_j^{n-1}),$$ where $w_j = e^{i\frac{2\pi j}{n}},$ the $n$-th roots of unity. Now, we just need to check none of the eigenvalues is zero: $$\mu_j = -\dfrac{2\cos(\frac{2\pi}{n})}{\sin^2(\frac{2\pi}{n})}+\dfrac{2\cos(\frac{2\pi j}{n})}{\sin^2(\frac{2\pi}{n})}.$$ Much to our dismay, this means that $\mu_1 = 0 $ or equivalently our Jacobian is singular at the point $(x^0, \lambda^0).$ This most likely means that either the Implicit Function Theorem is not strong enough, or there are more than one local extrema that satisfies that Lagrange multiplier equation. Either way, this remains an interesting problem...

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.