在△ABC中,求证sin(A+B)/(sinA+sinB)+sin(B+C)/(sinB+sinC)+sin(C+A)/(sinC+sinA)>=3/2

问题描述:

在△ABC中,求证sin(A+B)/(sinA+sinB)+sin(B+C)/(sinB+sinC)+sin(C+A)/(sinC+sinA)>=3/2

证明:先用正弦定理.将角度化成边:sin(A+B)/(sinA+sinB)+sin(B+C)/(sinB+sinC)+sin(C+A)/(sinC+sinA)=sinc/(sinA+sinB)+sina/(sinB+sinC)+sinb/(sinC+sinA)=1/(sina/sinc)+(sinb/sinc)+1/(sinb/sina)+(sinc/sina)+1...