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.


This question has an open bounty worth +150 reputation from dezdichado ending tomorrow.

Looking for an answer drawing from credible and/or official sources.

  • $\begingroup$ Is this a proposed problem? Or are you just speculating? $\endgroup$ – dezdichado Jun 14 at 15:25
  • $\begingroup$ Is there an issue if $n = 1$? $\endgroup$ – Gregory Jun 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 at 19:53

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...

  • $\begingroup$ +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 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 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 at 20:22

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.


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}

  • $\begingroup$ 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 at 19:45

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.

  • $\begingroup$ 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 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 at 23:43

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?

  • $\begingroup$ 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 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 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 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 at 19:49
  • $\begingroup$ Updated to use the Mean of the $x_i$ so that the property holds.... $\endgroup$ – Gopal Anantharaman Jun 20 at 16:48

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