Question

在三角形ABC中,sinA·sinB=sinC的平方-sinA的平方-sinB的平方，求角C

Answer 1

简单
sina/(sinb+sinc)+sinb/(sinc+sina)+sinc/(sina+sinb)
>sina/(sina+sinb+sinc)+sinb/(sinb+sinc+sina)+sinc/(sinc+sina+sinb)=1
下面证
sina<(sinb+sinc)
sin（b+c）=sinbcosc+sinccosb<(sinb+sinc)
即sinb（cosc-1）+sinc（cosb-1）<0
所以sina/(sinb+sinc)<1
下面用一个结论a>0
b>0
a>b
如果t>0
那么(a/b)<(a+t)/(b+t)展开既有
所以sina/(sinb+sinc)+sinb/(sinc+sina)+sinc/(sina+sinb)
<（sina+sina）/(sina+sinb+sinc)+（sinb+sinb)/(sinb+sinc+sina)+（sinc+sinc）/(sinc+sina+sinb)=2得证