sinA/(sinB+sinC)+sinB/(sinC+sinA) + sinC/(sinA+sinB)
=1/(sinB/sinA)+(sinC/sinA)+1/(sinC/sinB)+(sinA/sinB) + 1/(sinA/sinC)+(sinB/sinC)
=c/(a+b)+a/(b+c)+b/(a+c)
因此只要证明：c/(a+b)+a/(b+c)+b/(a+c)＜2即可!
a/(b+c)+b/(a+c)+c/(a+b) -2
=[a(a+c)(a+b)+b(b+c)(a+b)+c(a+c)(b+c)]-2(a+b)(a+c)(b+c)
=（a^3 +a^2b+a^2c+b^3+ab^2+cb^2 +c^3+c^2a+c^2b+3abc）-2(a^2b+a^2c+2abc+ac^2+b^2a +b^2c+bc^2)
=a^3+b^3+c^3-a^2b-a^2c-ab^2-cb^2-ac^2-bc^2- abc
=a^2［a-(b+c)］+ b^2［b- (a+c)］+c^2［c-(a+b)］- abc
＜0- abc=- abc……(∵b+c＞a,∴［a-(b+c) ＜0］,同理［b- (a+c)］＜0,［c-(a+b)］＜0)
∵a,b,c均为正数∴- abc＜0
故a/(b+c)+b/(a+c)+c/(a+b) -2＜0
∴c/(a+b)+a/(b+c)+b/(a+c)＜2
综上：sinA/(sinB+sinC)+sinB/(sinC+sinA) + sinC/(sinA+sinB) ＜2