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已知a+b+c=1,求证ab+bc+ca<=1/3
人气:304 ℃ 时间:2020-02-05 10:13:14
解答
1=(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac=1/2(a^2+b^2)+1/2(b^2+c^2)+1/2(a^2+c^2)+2ab+2bc+2ac>=ab+bc+ac+2ab+2bc+2ac=3ab+3bc+3ac=3(ab+bc+ac)
故而ab+bc+ca<=1/3
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