在直角三角型ABC中，斜边ＢＣ＝１，ＡＤ为ＢＣ上的高，求ＡＤ４次方／ＡＢ４次方＋２ＡＤ４次方／ＡＣ４次方的最小值为？ＢＣ／ＤＣ＝？
热心网友
1.解：AD^4/AB^4+2AD^4/AC^4=sinB^4+2sinC^4=cosC^4+2sinC^4=cosC^4+sinC^4+sinC^4=(cosC^2+sinC^2)^2-2sinC^2*(cosC^2)+sinC^4=1-2sinC^2*(1-sinC^2)+sinC^4=1-2sinC^2+2sinC^4+sinC^4=3sinC^4-2sinC^2+1=3[sicC^2-(1/3)}^2+(2/3)≥(2/3) 当且仅当sicC=√(1/3)时“＝”号成立－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－BC=AC/cosC；DC=AC*cosCBC/DC=(AC/cosC)/(AC*cosC)=1/cosC^2=1/(1/3)=3