文献[1]中提出了如下一个猜想不等式,此不等式结构优美、简单。猜想:若a,b,c∈R+,且a+b+c=3,则有(a~2)/(b+c)+(b~2)/(c+a)+(c~2)/(a+b)≥(a~2+b~2+c~2)/2。文献[2]给出上述不等式的证明,但通过构造函数和导数方法来证明,一般不易想到。经过研究,笔者提供几种应用人教A版《数学》(选修4-5)介绍的不等式方法来谈谈对上述不等式的简洁证法,以飨读者。证法1:应用排序不等式,不妨设a≥b≥c,则a+ The following conjecture inequality is proposed in [1]. This inequality structure is beautiful and simple. Conjecture: If a,b,c∈R+, and a+b+c=3, there are (a~2)/(b+c)+(b~2)/(c+a)+(c~2 )/(a+b)≥(a~2+b~2+c~2)/2. The literature [2] gives the above inequality proof, but it is not easy to think through the constructor function and the derivative method. After research, the author provides several kinds of inequality methods introduced by people who teach A version of “Mathematics” (Electives 4-5) to talk about the concise proof of the above inequality to readers. Proof 1: Applying a sorting inequality, you may wish to set a≥b≥c, then a+