xyz(x+y+z)=1.x(x+y+z)=1/(yz).x²+(y+z)x=1/(yz).x²+(y+z)x+(yz)=(yz)+[1/(yz)](x+y)(x+z)=(yz)+[1/(yz)]由基本不等式可知：(yz)+[1/(yz)]≥2.即(x+y)(x+z)≥2.∴[(x+y)(x+z)]min=2.