x,y,z∈(0,1),且x+y+z=2,求证1

Question

x,y,z∈(0,1),且x+y+z=2,求证1

Answer 1

先证右边.
易知任意实数x,y,z都有x²+y²+z²≥xy+yz+zx.
(x+y+z)²=x²+y²+z²+2(xy+yz+zx)≥3(xy+yz+zx)
∴xy+yz+zx≤(x+y+z)²/3=4/3.
再证左边.
xy+yz+zx
=xy+(x+y)(2-x-y)
=-x²+2x-xy+2y-y²
=-(1-x)²+y(1-x)+y-y²+1
=(1-x)(x+y-1)+y(1-y)+1
由已知z=2-x-y∈(0,1),得x+y-1>0,1-x>0,1-y>0,y>0,于是xy+yz+zx>1.
命题得证.