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
答
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...