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

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Uniteasy Admin · Accepted 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

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