tanA+tanB+tanC
=tan(A+B)(1-tanAtanB)+tanC
=tan(pai-c)(1-tanAtanB)+tanC
=-tanC(1-tanAtanB)+tanC
=tanAtanBtanC
所以:tanAtanBtanC=tanA+tanB+tanC
由题意,4tanBtanB=4tanAtanC
≤(tanA+tanC)^2
=(tanAtanBtanC-tanB)^2
=(tanBtanBtanB-tanB)^2
故4≤(tanB^2-1)^2,所以tanB大于等于根号3 所以B∏/3≤B