本刊1992年第六期《“配偶”技巧的应用》一文中的例1如下,设x、y、z是正数,求证: (y~2-x~2)/(z+x)+(z~2-y~2)/(x+y)+(x~2-z~2)/(y+z)≥0 (1) 上述这个不等式的背景实际上称作W Janous猜想,在《数学通讯》1992年第4期上作为数学难题刊载出来,据供题人称,此题选自《Crux》1612(即加拿大《数学难题》杂志) 由本刊及《数学通讯》上给出的两种证法实际上是同一种方法,显得繁琐。由此,我们不禁要问,对于一些分式不等式的证明,是否一定要实施通分或用排序原理等手段来完成呢?有无更好的一些方法?回答是肯定的。本文拟介绍一种较为简洁而又实用的分母置换证法,以供教学中参考。我们先以证明W.Janous猜测为例。如若 The 1st example of the article “Application of ”Spouse“ Technique” in the sixth issue of 1992 is as follows. Let x, y and z be positive numbers, and verify that: (y~2-x~2)/(z+x)+ ( z~2-y~2)/(x+y)+(x~2-z~2)/(y+z)≥0 (1) The background of the above inequality is actually called the W Janous conjecture. The Mathematical Newsletter, published in the fourth issue of 1992, was published as a math puzzle. According to the confessors, this question was selected from the “Crux” 1612 (Canada’s “Mathematics Puzzle” magazine). This method of authentication is actually the same method and it is tedious. From this, we cannot help but ask, for the proof of some fractional inequalities, whether it must be implemented through points or by using sorting principles to complete it? Are there any better ways? The answer is yes. This article will introduce a relatively simple and practical denominator replacement certificate method for reference in teaching. Let us first prove the case of W. Janous’s guess. If