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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.

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  • $\begingroup$ Is this a proposed problem? Or are you just speculating? $\endgroup$ – dezdichado Jun 14 '19 at 15:25
  • $\begingroup$ Is there an issue if $n = 1$? $\endgroup$ – Gregory Jun 19 '19 at 14:32
  • $\begingroup$ It does not really work if $n = 1$ or $2$ because in these cases the denominators can become $0$. I think it is well-defined for $n \geq 3$. $\endgroup$ – Tob Ernack Jun 20 '19 at 19:53
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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}

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  • $\begingroup$ This is interesting +1. Since the general Jacobian matrix I computed was singular, maybe there are multiple points where the inequalities are attained. $\endgroup$ – dezdichado Jun 19 '19 at 19:45
  • $\begingroup$ When $n = 4,$ it must be that $2x_1+x_2 < \pi$ with your choice. Have you actually checked that there are still infinitely many solutions satisfying that quadratic $\cot$ equation in terms of $x_1, x_2 ? $ $\endgroup$ – dezdichado Jun 27 '19 at 0:52
  • $\begingroup$ Yes, I checked $2x_1+x_2 < \pi$ for all $x_2 \in (0, \frac{\pi}{2}$ and corresponding $x_1 = \mathrm{arccot}\frac{\sqrt{2}-\cos x_2}{\sin x_2} $. I use Maple. $\endgroup$ – River Li Jun 27 '19 at 2:36
  • $\begingroup$ damn, that just means this problem is truly something else assuming that it is true. $\endgroup$ – dezdichado Jun 27 '19 at 7:15
  • $\begingroup$ However, I only prove that case $n=4,5$. I do not know the case $n > 5$. $\endgroup$ – River Li Jun 27 '19 at 8:59
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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...

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  • $\begingroup$ +1 I had a similar idea of using Lagrange multipliers but could not get past the computation of the determinant of that "almost-diagonal" matrix. $\endgroup$ – Tob Ernack Jun 20 '19 at 19:24
  • $\begingroup$ By the way, would the Implicit Function Theorem have been strong enough to prove global uniqueness, assuming the eigenvalues of the Jacobian were all nonzero? From what I recall, this would tell you that the point is a locally unique solution, but not necessarily globally. $\endgroup$ – Tob Ernack Jun 20 '19 at 19:43
  • $\begingroup$ @TobErnack, you are right. In general, if one could prove that the Jacobian is non-singular at all points in the domain, then that would have been enough. However, as the other answer established, there are infinitely many local extrema and that's what's very interesting about this problem. $\endgroup$ – dezdichado Jun 20 '19 at 20:22
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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.

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  • $\begingroup$ Do you have a reference for the multivariable Jensen's inequality? The only one I know are measure - theoretic versions and I am not sure if they apply here. $\endgroup$ – dezdichado Jun 13 '19 at 23:15
  • $\begingroup$ @dezdichado Unfortunately I am not certain. I found this reference but I cannot vouch for its correctness. In fact I am not sure the approach I used here is good at all, because it does not imply the existence of counterexamples, and also does not allow proving the truth of the claim, due to the Hessian matrix not being negative-definite. I'll leave the answer to give ideas to others that see why this approach fails. $\endgroup$ – Tob Ernack Jun 13 '19 at 23:43
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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?

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  • $\begingroup$ How does the 'Now ...' part follow? If $n$ is odd and $x_{\max}$ happens to be close to $\pi$, which is quite possible by considering the scenario such as $x_1 = \pi-(n-1)\epsilon$ and $x_2 = \cdots = x_n = \epsilon$ for sufficiently small $\epsilon$, then your bound $\sin^n(x_{\max})/\sin^n(2x_{\max})$ is actually negative, which is of course absurd. $\endgroup$ – Sangchul Lee Jun 19 '19 at 19:12
  • $\begingroup$ Saw from an answer above that, "First note that since $x_i>0$ and $x_1+x_2+⋯+x_n=\pi$, we must have $0<x_i<\pi$ and $0<x_{i+1}+x_i<\pi$ for all $i=1,…n$. Therefore we also have $\sin{x_i}>0$ and $\sin(x_{i+1}+x_i)>0$ so each term is strictly positive". $\endgroup$ – Gopal Anantharaman Jun 19 '19 at 19:40
  • $\begingroup$ I have no objection for that. But when replacing all $x_i$'s by $x_{\max}$, you can make the sum $x_{\max}+x_{\max}$ exceed $\pi$. $\endgroup$ – Sangchul Lee Jun 19 '19 at 19:43
  • $\begingroup$ So we select not an $x_{max}$ but some $x_k$ among the given $x_i$s s.t the condition "Now..." part follows. I don't quite know if it always exists $\endgroup$ – Gopal Anantharaman Jun 19 '19 at 19:49
  • $\begingroup$ Updated to use the Mean of the $x_i$ so that the property holds.... $\endgroup$ – Gopal Anantharaman Jun 20 '19 at 16:48
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I have what amounts to almost a proof. There is one key point that I cannot prove, so I've essentially just reduced this problem to that one.

Let $LHS(n)$ and $RHS(n)$ informally represent the left hand side and right hand side of the inequality in the case of n terms, respectively. This is simply done for brevity, and is meant to represent whichever form of the left hand side or right hand side is most sensible in the context.

We will prove the inequality by induction on the number of terms in the sequence $x_i$. First we prove the base case $n=3$. We have

$$ \begin{align} \frac{\sin x_1 \sin x_2 \sin x_3}{\sin(x_1 + x_2) \sin (x_2 + x_3) \sin (x_3 + x_1)} &= \frac{\sin x_1 \sin x_2 \sin x_3}{\sin(\pi - x_3)\sin(\pi - x_1)\sin(\pi - x_2)} \\ &= \frac{\sin x_1 \sin x_2 \sin x_3}{\sin(x_3)\sin(x_1)\sin(x_2)} \\ &= 1 \\ &= \Bigg(\frac{\sin \frac{\pi}{3}}{\sin \frac{2\pi}{3}} \Bigg)^3 \end{align} $$

Assume the statement holds for all sequences of $n$ or fewer terms. WLOG, since the problem is symmetric in the $x_i$, we can consider only ordered sequences with $x_1$ as the lowest term and $x_n$ as the highest. As pointed out by Martin R, we cannot order the sequence WLOG as the LHS is only invariant under cyclic permutations. However, the ordering was actually only necessary for failed attempts to show that $\phi \le \delta$ (see below), so the rest of the proof should be unaffected by this change.

For any sequence ${x_i}$ of $n+1$ terms, we can construct a sequence of $n$ terms $x_1, x_2, \cdots, x_{n-1}, (x_n + x_{n+1})$ by just combining the last two terms. By the induction assumption, this sequence satisfies the inequality: $LHS(n) \le RHS(n)$. Next, note that if we have $a \le b$, then we have $a\phi \le \delta b$ iff $\phi \le \delta$. If we can find $\phi \le \delta$ such that $LHS(n+1) = \phi LHS(n)$ and $\delta RHS(n) = RHS(n+1)$, then we will have $LHS(n+1) = \phi LHS(n) \le \delta RHS(n) = RHS(n+1)$.

We address the right side first.

$$ \begin{align} \delta\Bigg(\frac{\sin(\frac{\pi}{n})}{\sin(\frac{2\pi}{n})}\Bigg)^n &= \Bigg(\frac{\sin(\frac{\pi}{n+1})}{\sin(\frac{2\pi}{n+1})}\Bigg) ^{n+1} \\ \frac{\delta}{(2\cos(\frac{\pi}{n}))^n} &= \frac{1}{(2\cos(\frac{\pi}{n+1}))^{n+1}} \\ \delta &= \frac{\cos^n (\frac{\pi}{n}) }{2 \cos^{n+1} (\frac{\pi}{n+1})} \end{align} $$

This is asymptotically $\frac{1}{2}$ as $n \rightarrow \infty$.

Now we address the left hand side. We have

$$ \begin{gather*} \frac{\sin x_1 \cdots \sin x_n \sin x_{n+1}}{\sin(x_1 + x_2)\cdots \sin(x_n + x_{n+1})\sin(x_{n+1} + x_1)} = \phi \frac{\sin x_1 \cdots \sin x_{n-1} \sin(x_n + x_{n+1})}{\sin(x_1 + x_2)\cdots\sin(x_{n-1} +(x_{n} + x_{n+1}))\sin((x_{n} + x_{n+1})+x_1)} \\ \\ \phi = \frac{\sin x_n \sin x_{n+1} \sin(x_{n-1} + x_n + x_{n+1})\sin(x_n + x_{n+1} + x_1)}{\sin(x_{n-1} + x_n) \sin(x_n + x_{n+1})^2 \sin(x_{n+1} + x_1)} \end{gather*} $$

From this, once we show that $\phi \le \delta$ it follows easily. For any sequence of $n+1$ terms $x_i$ we have found $\phi$ and $\delta$, $\phi \le \delta$, such that $LHS(n+1) = \phi LHS(n)$ and $RHS(n+1) = \delta RHS(n)$. Thus, using the inductive assumption that $LHS(n) \le RHS(n)$, we have (as above) $LHS(n+1) \le RHS(n+1)$.

This is now the hiccup that I cannot prove. We must prove that $\phi \le \delta$. I have tinkered with sliders in desmos for $n=3$ (so $n+1=4$ terms), and it does seem to hold with a rather good margin. I have tried using the (now invalid) orderedness of the sequence to reduce $\phi$ to a reasonable expression. I have also tried splitting $\phi$ into a product of $4$ simpler fractions, then finding the maximum value for any sequence for each simpler fraction and using $\phi = abcd \le \max(abcd) \le \max(a)\max(b)\max(c)\max(d)$. However, this method proved ineffective, as I could not find an arrangement of the terms which did not have one of the simpler fractions being unbounded for some sequence.

Not a proof, but certainly progress. I'll update if I make any more progress on showing $\phi \le \delta$.

EDIT: Turns out $\phi$ is not strong enough. For $n=4$, taking the new sequence $(x_1, x_2, x_3, x_4, x_5)=(0.25,0.3416,0.5,1,1.05)$ gives $\phi = 0.401 > 0.361 = \delta$. However, the inequality still holds here - it's just that the RHS was larger than the LHS by enough of a margin to decrease by a larger factor and still hold. It may be possible to account for this initial difference, but it will not be easy by any means.

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  • $\begingroup$ Why is the problem symmetric in the $x_i$? Unless I am mistaken, the LHS is not invariant under arbitrary permutations of the $x_i$, only under cyclic permutations. $\endgroup$ – Martin R Jun 25 '19 at 14:37
  • $\begingroup$ @MartinR You're right, I hadn't thought about that too much. Luckily, the order isn't actually used for anything in the almost-complete proof, and was only necessary for an attempt to fix the hole, which failed anyways. $\endgroup$ – Vedvart1 Jun 25 '19 at 14:55
  • $\begingroup$ I thought about this approach too. But as other answers suggested, there seem to be many different points where the maximum is achieved. Thus, if the inequality is true, then none of the standard techniques of solving olympiad inequality should be sufficient as they only address inequalities whose equilibrium points are all equal, or most of them are equal. $\endgroup$ – dezdichado Jun 25 '19 at 21:12
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Note: this a trial approach via the Gamma function. Looks promising but at the moment I cannot find a way to close the last step.
Maybe someone can help to conclude ?

Since we are given $$ \left\{ \matrix{ 0 < x_{\,k} < \pi \hfill \cr x_{\,1} + x_{\,2} + \cdots + x_{\,n} = \pi \hfill \cr {{\sin x_{\,1} \sin x_{\,2} \cdots \sin x_{\,n} } \over {\sin \left( {x_{\,1} + x_{\,2} } \right)\sin \left( {x_{\,1} + x_{\,2} } \right) \cdots \sin \left( {x_{\,n} + x_{\,1} } \right)}} \le {{\sin \left( {\pi /n} \right)^{\,n} } \over {\sin \left( {2\pi /n} \right)^{\,n} }} \hfill \cr} \right. $$ and since it is $$ \Gamma \left( z \right)\,\Gamma \left( {1 - z} \right) = {\pi \over {\sin \left( {\pi \,z} \right)}}\quad \left| {\;\forall z \in \mathbb C\backslash \mathbb Z} \right. $$

Then, putting $$ x_{\,k} = \pi \,z_{\,k} \quad \left| {\;0 < z_{\,k} < 1} \right. $$ so $z_k \notin \mathbb Z$ and we can apply the relation with Gamma to obtain $$ \eqalign{ & {{\sin x_{\,1} \sin x_{\,2} \cdots \sin x_{\,n} } \over {\sin \left( {x_{\,1} + x_{\,2} } \right)\sin \left( {x_{\,1} + x_{\,2} } \right) \cdots \sin \left( {x_{\,n} + x_{\,1} } \right)}} = \cr & = {{\Gamma \left( {z_{\,1} + z_{\,2} } \right)\Gamma \left( {z_{\,2} + z_{\,3} } \right) \cdots \Gamma \left( {z_{\,n} + z_{\,1} } \right)} \over {\Gamma \left( {z_{\,1} } \right)\Gamma \left( {z_{\,2} } \right) \cdots \Gamma \left( {z_{\,n} } \right)}}\; \cdot \cr & \cdot \;{{\Gamma \left( {1 - z_{\,1} - z_{\,2} } \right)\Gamma \left( {1 - z_{\,2} - z_{\,3} } \right) \cdots \Gamma \left( {1 - z_{\,n} - z_{\,1} } \right)} \over {\Gamma \left( {1 - z_{\,1} } \right)\Gamma \left( {1 - z_{\,2} } \right) \cdots \Gamma \left( {1 - z_{\,n} } \right)}} = \cr & = {{z_{\,1} ^{\,\overline {\,z_{\,2} \,} } z_{\,2} ^{\,\overline {\,z_{\,3} \,} } \cdots z_{\,n} ^{\,\overline {\,z_{\,1} \,} } } \over {\left( {1 - z_{\,1} - z_{\,2} } \right)^{\,\overline {\,z_{\,2} \,} } \left( {1 - z_{\,2} - z_{\,3} } \right)^{\,\overline {\,z_{\,3} \,} } \cdots \left( {1 - z_{\,n} - z_{\,1} } \right)^{\,\overline {\,z_{\,1} \,} } }} = \cr & = {{z_{\,1} ^{\,\overline {\,z_{\,2} \,} } z_{\,2} ^{\,\overline {\,z_{\,3} \,} } \cdots z_{\,n} ^{\,\overline {\,z_{\,1} \,} } } \over {\left( { - z_{\,1} } \right)^{\,\underline {\,\,z_{\,2} \,} } \left( { - z_{\,2} } \right)^{\,\underline {\,\,z_{\,2} \,} } \cdots \left( { - z_{\,n} } \right)^{\,\underline {\,\,z_{\,2} \,} } }} \cr} $$ where $z^{\,\underline {\,w\,} } ,\quad z^{\,\overline {\,w\,} } $ represent respectively the Falling and Rising Factorial

It is in fact known that $$ \left( { - z} \right)^{\;\underline {\,w\,} } = {{\,\sin \left( {\pi \,\left( {z + w} \right)} \right)} \over {\sin \left( {\pi \,z} \right)\,}}\;z^{\,\overline {\,w\,} } $$ which corresponds to the "upper negation" identity for complex Binomials.

However, for our scope, we have better and restart from the second line and use the fact that $ln \Gamma(z)$ is convex, and decreasing in $(0,1)$.

That means that the average of the log of gamma at a number of points is no less than the log of gamma taken at the average point, or $$ \Gamma \left( {{1 \over n}} \right)^{\,n} \le \Gamma \left( {z_{\,1} } \right)\Gamma \left( {z_{\,2} } \right) \cdots \Gamma \left( {z_{\,n} } \right) $$ and consequently $$ \eqalign{ & \Gamma \left( {1 - {1 \over n}} \right)^{\,n} \le \Gamma \left( {1 - z_{\,1} } \right)\Gamma \left( {1 - z_{\,2} } \right) \cdots \Gamma \left( {1 - z_{\,n} } \right) \cr & \Gamma \left( {{2 \over n}} \right)^{\,n} \le \Gamma \left( {z_{\,1} + z_{\,2} } \right)\Gamma \left( {z_{\,2} + z_{\,3} } \right) \cdots \Gamma \left( {z_{\,n} + z_{\,1} } \right) \cr & \Gamma \left( {1 - {2 \over n}} \right)^{\,n} \le \Gamma \left( {1 - z_{\,1} - z_{\,2} } \right)\Gamma \left( {1 - z_{\,2} - z_{\,3} } \right) \cdots \Gamma \left( {1 - z_{\,n} - z_{\,1} } \right) \cr} $$

Therefore $$ \eqalign{ & \Gamma \left( {{2 \over n}} \right)^{\,n} \Gamma \left( {1 - {2 \over n}} \right)^{\,n} = \left( {{\pi \over {\sin \left( {\pi \,2/n} \right)}}} \right)^{\,n} \le \cr & \le \Gamma \left( {z_{\,1} + z_{\,2} } \right)\Gamma \left( {z_{\,2} + z_{\,3} } \right) \cdots \Gamma \left( {z_{\,n} + z_{\,1} } \right)\; \cdot \cr & \cdot \;\Gamma \left( {1 - z_{\,1} - z_{\,2} } \right)\Gamma \left( {1 - z_{\,2} - z_{\,3} } \right) \cdots \Gamma \left( {1 - z_{\,n} - z_{\,1} } \right) \cr & \Gamma \left( {{1 \over n}} \right)^{\,n} \Gamma \left( {1 - {1 \over n}} \right)^{\,n} = \left( {{\pi \over {\sin \left( {\pi \,/n} \right)}}} \right)^{\,n} \le \cr & \le \Gamma \left( {z_{\,1} } \right)\Gamma \left( {z_{\,2} } \right) \cdots \Gamma \left( {z_{\,n} } \right)\; \cdot \Gamma \left( {1 - z_{\,1} } \right) \Gamma \left( {1 - z_{\,2} } \right) \cdots \Gamma \left( {1 - z_{\,n} } \right) \cr} $$

It remains to demonstrate that by dividing the two we get the correct inequality,
i.e. that it is $$ \eqalign{ & 1 \le {{\left( \matrix{ \Gamma \left( {z_{\,1} + z_{\,2} } \right)\Gamma \left( {z_{\,2} + z_{\,3} } \right) \cdots \Gamma \left( {z_{\,n} + z_{\,1} } \right)\; \cdot \hfill \cr \cdot \;\Gamma \left( {1 - z_{\,1} - z_{\,2} } \right)\Gamma \left( {1 - z_{\,2} - z_{\,3} } \right) \cdots \Gamma \left( {1 - z_{\,n} - z_{\,1} } \right) \hfill \cr} \right)} \over {\left( {{\pi \over {\sin \left( {\pi \,2/n} \right)}}} \right)^{\,n} }} \le \cr & \le {{\Gamma \left( {z_{\,1} } \right)\Gamma \left( {z_{\,2} } \right) \cdots \Gamma \left( {z_{\,n} } \right)\; \cdot \Gamma \left( {1 - z_{\,1} } \right) \Gamma \left( {1 - z_{\,2} } \right) \cdots \Gamma \left( {1 - z_{\,n} } \right)} \over {\left( {{\pi \over {\sin \left( {\pi \,/n} \right)}}} \right)^{\,n} }} \cr} $$

That reads that the average of $\ln \Gamma(z_k + z_{k+1})$ is less dispersed around $\ln \Gamma(z_{2\, avg})$ than $\ln \Gamma(z_k)$ around $\ln \Gamma(z_{1\, avg})$,
which intuitively makes sense (for large $n$ at least).

It remains to express that in a formal way.

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If $x1+x2+x3+x4=\pi$ then,

$$4 \sin (\text{x1}) \sin (\text{x2}) \sin (\text{x3}) \sin (\text{x4})-\sin (\text{x1}+\text{x2}) \sin (\text{x1}+\text{x4}) \sin (\text{x2}+\text{x3}) \sin (\text{x3}+\text{x4}) = -(\sin (\text{x1}) \sin (\text{x3})-\sin (\text{x2}) \sin (\text{x4}))^2.$$

Which proves the n=4 case and also shows that the points of equality are given by

$$\sin (\text{x1}) \sin (\text{x3})=\sin (\text{x2}) \sin (\text{x4}).$$

You can prove the identity in quite a long winded way by expressing everything on the LHS as cosines in some combination of x1, x2, x3 after setting x4 to be $\pi-x1-x2-x3$. Most terms cancel and you get

$$-(\cos (\text{x1}+2 \text{x2}+\text{x3})+\cos (\text{x1}-\text{x3})-2 \cos (\text{x1}+\text{x3}))^2$$

which simplifies to give the RHS.

Perhaps one can generalise this identity to prove the other cases?

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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 $x_i>0$ 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 $\frac{\pi}{n}-q\epsilon>0$

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 :

$\exists \,\epsilon>0$ 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 $0<b_i=x_i+x_{i+1}\leq \frac{\pi}{2}$ and $0<a_i=x_i<\frac{\pi}{2}$

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

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  • $\begingroup$ @RiverLi can you confirm there is no mistakes?Thanks! $\endgroup$ – Erik Satie May 4 at 9:35
  • $\begingroup$ You investigated the special case $|x_i-x_{i+1}|=\epsilon$ for all $i$. Have you done any other cases? $\endgroup$ – River Li May 6 at 1:38
  • $\begingroup$ @RiverLi no but the second idea was to perturb (I don't know if it is the good word?) the variables two by two and see what happend. $\endgroup$ – Erik Satie May 6 at 8:28
  • $\begingroup$ @RiverLi Furthermore we can apply Popoviciu's inequality (a refinement of Jensen's inequality) to get a variant of the inequality but I don't know if there are counter-examples .Good day . $\endgroup$ – Erik Satie May 6 at 8:31
  • $\begingroup$ @RiverLi I have good news see my "another way''.Good day . $\endgroup$ – Erik Satie May 18 at 12:05

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