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求证三角形ABC,1/(1+cosA+cosB)+1/(1+cosB+cosC)+1/(1+cosC+cosA)>=2
是1/(1+cos^2A+cos^2B)+1/(1+cos^2B+cos^2C)+1/(1+cos^2C+cos^2A)
人气:468 ℃ 时间:2019-11-11 12:47:47
解答
证明:cosA+cosB=2cos[(A+B)/2]*cos[(A-B)/2] =2sin(C/2)cos[(A-B)/2] ≤2sin(C/2)cosA+cosB =(b²+c²-a²)/2bc+ (a²+c²-b²)/2ac =(ab²+ac²+a²b+bc²-a³-b³)/2abc =(a+b)(2ab+c²-a²-b²)/2abc ≤(a+b)c/2abcos²A+cos²B =1+(cos2A+cos2B)/2 =1+cos(A+B)cos(A-B) =1-cosCcos(A-B) ≥1-cosC1/(1+cos²A+cos²B)+1/(1+cos²B+ cos²C)+1/(1+cos²C+cos²A)≤2 ≤1/(2-cosC)+1/(2-cosA)+1/(2-cosB) 不等式当且仅当A=B=C=60°时,取等号
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