问题标题：
【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
陈凯回答：
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...
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