证明：法一:x(1-y)+y(1-z)+z(1-x) =(1-y-z)x+y(1-z)+z1)若02)若y+z>1,(1-y-z)x+y(1-z)+z =(y-1)(1-z)+1<1得证法二:构造函数f(x)=(1-y-z)x+y(1-z)+z其中f(x)是关于x的一次函数,0因为f(0)=y+z-yz-1=(1-y)(z-1)<0 f(1)=(1-y-z)+y+z-yz-1=-yz<0所以对于0所以(1-y-z)x+y(1-z)+z《0即x(1-y)+y(1-z)+z(1-x)＜1
