Question

求证：在△ABC中,1<sinA/(sinB+sinC)+sinB/(sinC+sinA)+sinC/(sinA+sinB)<2

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