XYZ(X+Y+Z)=1 求（x+y)(x+z)的最小值（X>0,Y>0,Z>O)
问题描述：
XYZ(X+Y+Z)=1 求（x+y)(x+z)的最小值（X>0,Y>0,Z>O)
答
XYZ(X+Y+Z)=X^2+XY+XZ=1/YZ
(X+Y)(X+Z)=X^2+XY+XZ+YZ=1/YZ+YZ=2
答
(x+y)(x+z)=x(x+y+z)+yz=1/yz+yz>=2
答
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.
答
XYZ(X+Y+Z)=(X2+XY+XZ)YZ=1 则X2+XY+XZ=1/YZ
所以（x+y)(x+z)=X2+XY+XZ+YZ=YZ+1/YZ
剩下的就不用说了吧