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最近本刊文载有如下一道1991年苏联农村中学生数学竞赛十一年级试题: 已知a,b,c>0,求证 (a+b)(b+c)(c+a)≥8abc (1) 对a、b、c每两个应用均值不等式,然后相乘即得(1)式。这道试题看似简单平常,但实质上却隐含着丰富的内容。很多数学竞赛题就是以它为源头,通过变换逐步演绎深化而成。真可谓是金线串珠,妙趣无穷。 1.对(1)作变换:a→b+c,b→c+a,c→a+b,则得现行教科书代数下册p32的一道复习参考题:已知a,b,c∈R~+,求证 2(a~3+b~3+c~3)≥a~2(b+c)+b~2(c+a)+c~2(a+b) (2)
The most recent article contains the following questions in the eleventh grade of the 1991 Soviet Rural Middle School Mathematics Contest: Known a,b,c>0, Proof (a+b)(b+c)(c+a)≥8abc (1 For each of a, b, and c, the mean inequality is applied and then multiplied to obtain (1). This test appears to be simple and ordinary, but in essence it implies rich content. Many mathematics competition questions take it as a source and gradually transform and deepen it. It can be said that it is gold threading, endless fun. 1. For (1) transformation: a → b + c, b → c + a, c → a + b, then get a review of the current textbook algebra under the book p32: known a, b, c ∈ R ~+, Proof 2(a~3+b~3+c~3)≥a~2(b+c)+b~2(c+a)+c~2(a+b) (2)