幾何寶庫

  設外接圓半徑為R、內切圓半徑為r、三邊長為a,b,c、△ABC面積為S r=2*S/(a+b+c)=(2*1/2*ab*sinC)/(2R*sinA+2R*sinB+2R*sinC) =(2R*sinA*2R*sinB*sinC)/(2R*sinA+2R*sinB+2R*sinC) =2R*sinA*sinB*sinC/(sinA+sinB+sinC) =1/2*R*sinA*sinB*sinC/(cosA/2*cosB/2*cosC/2) =4R*sinA/2*sinB/2*sinC/2 ∴AI=r/(sinA/2)=4R*sinB/2*sinC/2 BI=r/(sinB/2)=4R*sinA/2*sinC/2 CI=r/(sinC/2)=4R*sinA/2*sinB/2 由Jesen不等式可得 3/4≧(sinA/2)^2+(sinB/2)^2+(sinB/2)^2 又 (sinA/2)^2+(sinB/2)^2+(sinB/2)^2≧sinA/2*sinB/2+sinB/2*sinC/2+sinC/2*sinA/2 故 3≧4sinA/2*sinB/2+4sinB/2*sinC/2+4sinC/2*sinA/2 →3R≧4R*sinA/2*sinB/2+4R*sinB/2*sinC/2+4R*sinC/2*sinA/2 →3R≧AI+BI+CI →AO+BO+CO≧AI+BI+CI 得證