《数学教学》1992年第5期《学会变更问题的形式》一文列举的例2及分析1如下: “例2 已知:a_1,a_2,…,a_n是n个正数,满足a_1a_2…a_n=1。求证:(2+a_1)(2+a_2)…(2+a_n)≥3~n(89年全国高中数学联赛试题)。分析1:昨一看,太简单了, 问题出在哪里?分析求证的结论和以上的证法,发现忽视了求证结论中2与3的矛盾,欲证的不等式的左边每一个因式都是两个正数的和,且有一个是2,而右边是3的n次方。有了,两个正数的和可以变为三个正数的和,即 The Example 2 and Analysis 1 listed in the article “Forms of Learning to Change the Problem” in Mathematics Teaching, 5th Issue, 1992 are as follows: “Example 2 Known: a_1, a_2,..., a_n are n positive numbers, satisfying a_1a_2...a_n= 1. Proof: (2+a_1)(2+a_2)...(2+a_n)≥3~n (National Mathematical League High School Test Question for 89 years) Analysis 1: Yesterday, it was too simple, where was the problem? Analysing the conclusions of the verification and the above proofs, we find that we neglected the contradiction between the 2 and 3 in the conclusion of the verification. The left side of the inequality to be proved is a sum of two positive numbers, and one is 2, and the right is The nth power of 3. With that, the sum of two positive numbers can become the sum of three positive numbers, ie