tanA,tanB,tanC都是正数,所以A,B,C都是锐角
lgtanA+lgtanC=2lgtanB
tanAtanC=tan²B
又tanA+tanB+tanC
=tan(A+B)(1-tanAtanB)+tanC
=tan(π-C)(1-tanAtanB)+tanC
=-tanC(1-tanAtanB)+tanC
=tanAtanBtanC
即tanAtanBtanC=tanA+tanB+tanC
4tan²B=4tanAtanC
≤(tanA+tanC)²
=(tanAtanBtanC-tanB)²
=(tanBtanBtanB-tanB)²
4≤(tan²B-1)²
tan²B-1≥2或tan²B-1≤-2
tan²B≥3或 tan²B ≤-1 (舍)
所以 tanB≥√3
所以 π/3≤B