Given positive numbers x , y , z , if xyz(x+y+z)=1 , find the minimum value of (x+y)(y+z) . Steven Zheng posted 3 months ago 0 0 0
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 Steven Zheng posted 3 months ago