=(x+y+z-z)(x+y+z-x)

=(x+y+z)^2-(x+z)(x+y+z)+xz

=(x+y+z)^2-(x+z+y-y)(x+y+z)+xz

=(x+y+z)^2-(x+y+z)^2+y(x+y+z)+xz

=y(x+y+z)+xz

Because xyz(x+y+z)= 1, so y(x+y+z)= 1/xz, so the original formula can be changed to xz+( 1/xz), obviously the minimum value is 2. (xz can be regarded as the square of xz under the root sign, and 1/xz is the square of 1/xz under the root sign, and the minimum value can be found by the formula as 2).