x.y.z都大于0,xy+yz+xz=1,求1/(x+y)+1/(x+z)+1/(y+z)的最小值

来源：学生作业帮助网编辑：六六作业网时间：2024/11/25 17:22:29

x.y.z都大于0,xy+yz+xz=1,求1/(x+y)+1/(x+z)+1/(y+z)的最小值x.y.z都大于0,xy+yz+xz=1,求1/(x+y)+1/(x+z)+1/(y+z)的最小值x.

x.y.z都大于0,xy+yz+xz=1,求1/(x+y)+1/(x+z)+1/(y+z)的最小值
x.y.z都大于0,xy+yz+xz=1,求1/(x+y)+1/(x+z)+1/(y+z)的最小值

x.y.z都大于0,xy+yz+xz=1,求1/(x+y)+1/(x+z)+1/(y+z)的最小值
将x y 设定为0:0.01:100（范围0-100,步长0.01）,用matlab算了一下,给出的最小结果是 0.044996,以下是Matlab程序,凑合着看看能用不.
for x = 0:0.01:100
for y = 0:0.01:100
z = (1-x.*y)./(x+y);
t = 1./(x+y) + 1./(x+z)+1./(y+z);
end
end
min(t)

x.y.z都大于0,xy+yz+xz=1,求1/(x+y)+1/(x+z)+1/(y+z)的最小值已知ix大于y大于z,且x+y+z=0,下列不等式中成立的是A、xy>yz B、xz>yz C、xy>xz D、x|y|＞z|y| 已知：X,Y,Z均大于0且小于1,X+Y+Z=2,W=XY+YZ+XZ,求W的取值范围? 证明当x+y+z=1时,x/yz+y/xz+z/xy≥9 xyz-xy-xz+x-yz+y+z-1因式分解还是因式分解 xyz-xy-xz+x-yz+y+z-1 X+y+z=1 xy+yz+xz=0 x^2+y^2+z^2=? 已知,xyz=0,求x/(xy+x+1)+y/(yz+y+1)+z/(xz+z+1）值? 已知x+y+z=1,xy+yz+xz=0,求x^2+y^2+z^2的值. 设x,y,z∈R+,xy+yz+xz=1,证明不等式:(xy)^2/z+(xz)^2/y+(yz)^2/x+6xyz≥x+y+zRt (2X+Z-Y)/(X^2-XY+XZ-YZ)-(Y-Z)/(X^2-XY-XZ+YZ) 已知xy(x+y)^-1=1,yz(y+z)^-1=2,xz(z+x)^-1=3,试求xyz(xy+yz+xz)^-1的值已知xy(x+y)^-1=1,yz(y+z)^-1=2,xz(z+x)^-1=3,试求xyz(xy+yz+xz)^-1的值 x+y+z=5,xy+xz+yz=1 ,求Z的最小值和最大值已知x,y,z是实数,且xyz=1,求证x^2+y^2+z^2+3大于等于2(xy+xz+yz) 已知x2+y2+z2-xy-yz-xz=0,求证x=y=z -x=3,/y/=4,z+3=0,求xy+yz+xz的值 z=f(x,y) xy+yz+xz=1 ,求dz