A+B=π-C
tan(A+B)=tan(π-C)
(tanA+tanB)/(1-tanAtanB)=(tanπ-tanC)/(1+tanπtanC)
整理可得
tanA+tanB+tanC=tanAtanBtanC
(tanAtanBtanC)^3=(tanA+tanB+tanC)^3>=27tanAtanBtanC
(tanAtanBtanC)^2>=27
tanAtanBtanC>=3√3
A+B=π-C
tan(A+B)=tan(π-C)
(tanA+tanB)/(1-tanAtanB)=(tanπ-tanC)/(1+tanπtanC)
整理可得
tanA+tanB+tanC=tanAtanBtanC
(tanAtanBtanC)^3=(tanA+tanB+tanC)^3>=27tanAtanBtanC
(tanAtanBtanC)^2>=27
tanAtanBtanC>=3√3