《数学通报》1999年第3期给出了这样一道不等式证明题:设a,b,c为正数,证明bc(b+c)+ca(c+a)+ab(a+b)　≤32(b+c)(c+a)(a+b).该刊第4期上的证明是通过构造三角形,证明原不等式等价于△ABC中的已知不等式sinA2+sinB2+sinC2≤32.下面,我们试给出原不等式的一种直接并且简单的证法. “Mathematical Bulletin”, No. 3, 1999, gives an inequality proof: Let a, b, and c be positive numbers, and prove that bc(b+c)+ca(c+a)+ab(a+b) ≤ 32(b+c)(c+a)(a+b). The proof in the fourth issue of this paper is to construct a triangle and prove that the original inequality is equivalent to the known inequality in △ABC sinA2+sinB2+sinC2≤32 Below, we try to give a direct and simple proof of the original inequality.