证明 设△PBC，△PCA，△PAB的面积分别为x/2，y/2，z/2，令BC=a，CA=b，AB=c。则有:a*PA≥y+z，b*PB≥z+x，c*PC≥x+y。a*PD=x，b*PE=y，c*PF=z。所以得:PA/(PA+kPD)≥(y+z)/(y+z+kx) ，PB/(PB+kPE)≥(z+x)/(z+x+ky) ，PC/(PC+kPF)≥(x+y)/(x+y+kz) 。故只需证:(y+z)/(y+z+kx)+(z+x)/(z+x+ky)+(x+y)/(x+y+kz)≥6/(2+k) 上式化简整理为:[k(y-x)+k(z-x)]/(y+z+kx)+[k(z-y)+k(x-y)]/(z+x+ky)+[k(x-z)+k(x-y)]/(x+y+kz)≥0<===>k(k-1)(y-z)^2/[(x+y+kz)(z+x+ky)]+k(k-1)(z-x)^2/[(y+z+kx)(x+y+kz)]+k(k-1)(x-y)^2/[(z+x+ky)(y+z+kx)] ≥0当k≥1时,上式显然成立,故不等式成立.