Question

在锐角三角形中,求证:tanA*tanB*tanC*>1

Answer 1

tanA=tan[180º-(B+C)]=-tan(B+C)=-(tanB+tanC)/(1-tanBtanC)=(tanB+tanC)/(tanBtanC-1).===>tanA(tanByanC-1)=tanB+tanC.===>tanAtanBtanC=tanA+tanB+tanC.因A,B,C均为锐角,故tanA,tanB,tanC＞0.由均值不等式可知,tanAtanBtanC=tanA+tanB+tanC≥3(tanAtanBtanC)^(1/3).===>(tanAtanBtanC)^(2/3)≥3.===>tanAtanBtanC≥3^(3/2)=3√3,即tanAtanBtanC≥3√3＞1.其中的等号仅当A=B=C=60º时取得.