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任意△ABC,求证:1/2(1/a+1/b+1/c)
人气:376 ℃ 时间:2020-06-16 06:40:43
解答
1)题是1/2*(1/a+1/b+1/c)=2ab,b^2+c^2>=2bc,c^2+a^2>=2ca a^2+b^2+c^2>=ab+bc+ca (cosA/a+cosB/b+cosC/c)-0.5(1/a+1/b+1/c) =(a^2+b^2+c^2)/(2abc)-(ab+bc+ca)/(2abc)>=0 cosA/a+cosB/b+cosC/c≥0.5(1/a+1/b+1/c) ...
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