数学不等式题:x.y.z属于R+,xyz(x+y+z)=1 求(x+y)(y+z)最小值
用均值不等式解题若x>0,y>0,z>0,且xyz(x+y+z)=1,求(x+y)(y+z)的最小值
x+y+z=1 求xyz/(x+y)(y+z)(z+x)的最大值
xyz∈R+且 x+2y+3z=36求 1/x +2/y +3/z的最小值
已知xyz属于R+,x+y+z=1,求证x^3/(y(1-y))+y^3/(z(1-z))+z^3/(x(1-x))大于
已知x,y,z属于R+(正实数),且xyz(x+y+z)=4+2*根号下3,则(x+y)(y+z)的最小值是?
已知x+y-z/z=x-y+z/y=-x+y+z/x,且xyz不等于0,求分式[(x+y)(x+z)(y+z)]/xyz
已知x,y,z属于正实数,且xyz(x+y+z)=1,则(x+y)(y+z)的最小值为?
若z分之x+y+z=y分之x-y+z=x分之-x+y+z,求xyz分之(x+y)(y+z)(z+x)
(x+y-z)/z=(y+z-x)/x=(z+x-y)/y 求(x+y)(y+z)(z+x)/xyz
已知:(x+y-z)/z=(x-y+z)/y+(y+z-x)/x,且xyz≠0,求代数式[(x+y)(y+z)(x+z)
已知x+y+z=xyz 求X^2+Y^2+Z^2+2/(XYZ)的最小值 谢谢了,原理上应该是柯西不等式题
已知x,y,z属于R+,x+y+z=3,(1)求1/x+1/y+1/z的最小值,(2)证明:3