证明在△ABC中.sinA/(sinB+sinC)+sinB/(sinC+sinA)+sinC/(sinA+sinB)<

1个回答

  • 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