2014年安徽省高中数学竞赛有如下一个不等式试题:已知正实数x,y,z满足x+y+z=1,求证:(z-y)/(x+2y)+(x-z)/(y+2z)+(y-x)/(z+2x)≥0.该题除标准答案提供的证法外,本文提供一些另外有趣的证明方式,同时给出不等式的一个推广.1.多种证法.证法一由x+y+z=1,可知原不等式等价于下述不等式 The 2014 Anhui High School Math Contest has the following inequality test: Known positive real numbers x, y, z satisfy x + y + z = 1, and prove that: (zy) / (x + 2y) + (xz) / (y + 2z ) + (yx) / (z + 2x) ≥ 0. In addition to the proofs provided by the standard answer, this article provides some other interesting proofs and a generalization of the inequalities. Law from x + y + z = 1, we can see that the original inequality is equivalent to the following inequality