高中不等式一题求解x,y,z都属于(0,1),求证x(1-y)+y(1-z)+z(1-x)
1个回答
证明:法一:
x(1-y)+y(1-z)+z(1-x)
=(1-y-z)x+y(1-z)+z
1)若0
相关问题
求解一道较难的不等式证明题目x,y,z∈[0,1] 求证(1+x)(1+Y)(1+Z)>=√8(x+y)(y+z)(x+
求一不等式证明.已知:x,y,z>=0,x+y+z=6.求证:(x+1/x)(y+1/y)(z+1/z)>=125/8这
已知x y z都属于R 2^x=3^y=6^z 求证 1/x+1/y=1/z
设0<x,y,z<1,求证x(1-y)+y(1-z)+z(1-x)<1
x,y,z,1/x+1/y+1/z=2,求证:√(x+y+z) ≥√(x-1)+ √(y-1)+ √(z-1)
x,y,z都>0.且x+y+z=1.求证1/x+4/x+9/z>=36
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
x,y,z都是正数,且xyz=1,求证:(1+x+y)(1+y+z)(1+z+z)≥27.
已知x+1/y=y+1/z=z+1/x,求证y/x+z/y+x/z=3
x,y,z属于R,且xyz(x+y+z)=1,求证(x+y)(y+z)≥2