Question

Given positive numbers x , y , z , if xyz(x+y+z)=1 , find the minimum value of (x+y)(y+z) .

Steven Zheng Steven Zheng posted 1 month ago


Answer

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 Steven Zheng posted 1 month ago

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