難度：☆☆☆☆ 題目： I,O分別為三角形的內心和外心 試證明AO+BO+CO≧AI+BI+CI 看解答請點此
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
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