常见一些中学数学杂志讨论下列不等式的证明:设A,B,C为三角形三内角,则 sinA+sinB+sinC≤3/2 3~(1/2)。但均限于运用三角函数之变形推出结论。本文拟用几何定理证明上述结论,并加以推广。我们先给出一个引理。引导在圆的内接n边形中,以内接正n边形之周长为最长。问题设A,B,C为三角形的三个内角,则 sinA+sinB+sinC≤(3/2)3~(1/2)。证明设α=ZA,β=2B,γ=2C 则α+β+γ=2(A+B+C)=2π Some secondary school mathematics magazines discuss the following inequalities: Let A, B, and C be the triangular inner angles, then sinA+sinB+sinC≤3/2 3~(1/2). However, they are limited to the use of trigonometric distortions. This article proposes to use the geometric theorem to prove the above conclusions and generalize it. Let us first give a lemma. Guided in the inscribed n-edge of the circle, the longest perimeter is the inner n-edge. The problem is that A, B, and C are three internal angles of a triangle, and sinA+sinB+sinC≤(3/2)3~(1/2). Prove that α = ZA, β = 2B, γ = 2C then α + β + γ = 2 (A + B + C) = 2π