Question
Given positive numbers x , y , z , if xyz(x+y+z)=1 , find the minimum value of (x+y)(y+z) .
Given positive numbers x , y , z , if xyz(x+y+z)=1 , find the minimum value of (x+y)(y+z) .
Distribute terms within the parentheses
And then use AM-GM Inequality
(x+y)(y+z)=xy+y2+yz+zx
=y(x+y+z)+zx≥2\sqrt{y(x+y+z)·zx}
=2
So the minimum value of (x+y)(y+z) is 2