Question

如果x+y+z=1/xyz那么 (x+y)(x+z)的最小值是多少

Answer 1

(x+y)(x+z)=x^2+xy+xz+yz=x^2+xy+z(x+y)=x^2+xy+z(1/xyz-z)=x^2+xy+1/xy-z^2

=xy+1/xy+(x+z)(x-z)=xy+1/xy+(1/xyz-y)(x-z)

=xy+1/xy+yz+1/yz-xy-1/xy

=yz+1/yz>=2

(x,y,z >0)