已知x,y,z为正实数，且 x+y+z<=3xyz 求1/1+x＋1/1+y＋1/1+z的值域

解：显然x，y，z→∞时1/1+x＋1/1+y＋1/1+z→最小值：0；
且显然有x+y+z<=3xyz
x,y,z为正实数
故有
（xyz）开3方《(x+y+z)/3《xyz
故xyz》1
故1/1+x＋1/1+y＋1/1+z
=[xy+yz+zx+2(x+y+z)+3]/(xyz+xy+yz+zx+x+y+z+1) (通分的结果）
=3[xy+yz+zx+2(x+y+z)+3]/[3xyz+3(xy+yz+zx)+3(x+y+z)+3]
《3[xy+yz+zx+2(x+y+z)+3]/[x+y+z+3(xy+yz+zx)+3(x+y+z)+3]
=3[xy+yz+zx+2(x+y+z)+3]/[2(xy+yz+zx)+4(x+y+z)+6+xy+yz+xz-3]
由xy+yz+zx》3（xyz）的2/3次方》3*1=3，
3[xy+yz+zx+2(x+y+z)+3]/[2(xy+yz+zx)+4(x+y+z)+6+xy+yz+xz-3]
《3[xy+yz+zx+2(x+y+z)+3]/[2(xy+yz+zx)+4(x+y+z)+6+3-3]
=3[xy+yz+zx+2(x+y+z)+3]/[2(xy+yz+zx)+4(x+y+z)+6]=3/2
故1/1+x＋1/1+y＋1/1+z的值域：(0,3/2]