西部数学奥赛一题的三角证法

• 作 者: 宿晓阳
• 机 构: 四川省成都实验外国语学校 (610031)
• 出 处: 中学生数学
• 发表日期: 2008年03期
• 关键词: 证法 正实数 均值不等式 李延林 下乙 石丁 二万

题目求证:对任意正实数a、b、c都有1≤(a/(a~2+b~2)~(1/2))+(b/(b~2+c~2)~(1/2))+(c/(c~2+a~2)~(1/2))≤(3 2~(1/2)/2).《中学生数学》2007年1月(上)P_(36)刊出了杨安琪同学对此题的“新解”.可惜原文的最后两步有误.因由cos~2α+cos~2β+cos~2γ≥3·(cos~2α·cos~2β·cos~2γ)~(1/3),cos~2αcos~2βcos~2γ≤(1/8),不能推出cos~2α+cos~2β+cos~2γ≥3·(1/8)~(1/3).尽管原文出现小小的失误,但原文的思想方法很好.

Question verification: For any positive real numbers a, b, c, there is 1 ≤(a/(a~2+b~2)~(1/2))+(b/(b~2+c~2)~( 1/2))+(c/(c~2+a~2)~(1/2))≤(3 2~(1/2)/2). “Middle School Mathematics” January 2007 (I) P_(36) published the “new solution” to this question by Yang Anqi. It is a pity that the last two steps of the original text were wrong. The cause was cos~2α+cos~2β+cos~2γ≥3·(cos~2α·cos ~2β·cos~2γ)~(1/3),cos~2αcos~2βcos~2γ≤(1/8), cannot be introduced cos~2α+cos~2β+cos~2γ≥3(1/8)~ (1/3). Although there was a slight mistake in the original text, the original text was very good.