A result of Vietoris states that if the real numbers a1,…,an\documentclass[12pt]{minimal}
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\begin{document}$$a_1,\ldots ,a_n$$\end{document} satisfy (*)a1≥a22≥⋯≥ann>0anda2k-1≥a2k(1≤k≤n/2),\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} \text{(*) } \qquad a_1\ge \frac{a_2}{2} \ge \cdots \ge \frac{a_n}{n}>0 \quad \text{ and } \quad a_{2k-1}\ge a_{2k} \quad (1\le k\le n/2), \end{aligned}$$\end{document}then, for x1,…,xm>0\documentclass[12pt]{minimal}
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\begin{document}$$x_1,\ldots ,x_m>0$$\end{document} with x1+⋯+xm<π\documentclass[12pt]{minimal}
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\begin{document}$$x_1+\cdots +x_m <\pi $$\end{document}, (**)∑k=1naksin(kx1)⋯sin(kxm)km>0.\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} \begin{aligned} \text{(**) } \qquad \sum _{k=1}^n a_k \frac{\sin (k x_1) \cdots \sin (k x_m)}{k^m}>0. \end{aligned} \end{aligned}$$\end{document}We prove that (∗∗)\documentclass[12pt]{minimal}
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\begin{document}$$(**)$$\end{document} (with “≥\documentclass[12pt]{minimal}
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\begin{document}$$\ge $$\end{document}” instead of “>”) holds under weaker conditions. It suffices to assume, instead of (∗)\documentclass[12pt]{minimal}
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\begin{document}$$(*)$$\end{document}, that ∑k=1Naksin(kt)k>0(N=1,…,n;0<t<π),\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} \sum _{k=1}^N a_k \frac{\sin (kt)}{k}>0 \quad (N=1,\ldots ,n; \, 0<t<\pi ), \end{aligned}$$\end{document}and, moreover, (∗∗)\documentclass[12pt]{minimal}
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\begin{document}$$(**)$$\end{document} is valid for a larger region, namely, x1,…,xm∈(0,π)\documentclass[12pt]{minimal}
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\begin{document}$$x_1,\ldots ,x_m\in (0,\pi )$$\end{document}.