由三角形ABC的面积为S＝〔（a＋b＋c）r〕／2＝(ab*sinC)/2,由正弦定理的,a/sinA=b/sinB=c/sinC=2R,则2Rr(sinA+sinB+sinC)/2=4R^2(sinA*sinB*sinC)/2,r/R=2(sinA*sinB*sinC)/(sinA+sinB+sinC) 那么下面证明2(sinA*sinB*sinC)/(sinA+sinB+sinC)=4sin(A/2)sin(B/2)sin(C/2)即可接下来，2(sinA*sinB*sinC)/(sinA+sinB+sinC)=4sin(A/2)sin(B/2)sin(C/2)这个式子要如何证明？