Question

求解：XYZ(X+Y+Z)=1 求（x+y)(x+z)的最小值（X>0,Y>0,Z>O)

Answer 1

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.

Answer 2

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

Answer 3

(x+y)(x+z)=x(x+y+z)+yz=1/yz+yz>=2

Answer 4

XYZ(X+Y+Z)=X^2+XY+XZ=1/YZ
(X+Y)(X+Z)=X^2+XY+XZ+YZ=1/YZ+YZ=2