已知三角形ABC的三边abc成等比数列,求证:aCos^2C/2+cCos^2A/2大于等于3/2b
问题描述：
已知三角形ABC的三边abc成等比数列,求证:aCos^2C/2+cCos^2A/2大于等于3/2b
答
证明：
ac=b²
∴b=√(ac)
左边=a(1+cosC)/2+c(1+cosA)/2
=(a²+b²+2ab-c²)/(4b)+(b²+c²+2bc-a²)/(4b)
=(2b²+2ab+2bc)/(4b)
=(b+a+c)/2
=b/2+(a+c)/2
≥b/2+√(ac)
=b/2+b
=3b/2
得证