Minimum value problem 
Find the minimum value of $(x+y)(y+z)$ where $x,y,z$ are positive real numbers satisfying the condition $$xyz(x+y+z)=1$$

Hint?

Just apply AM-GM.
$(x+y)(y+z)=xy+y^2+xz+yz=y(x+y+z)+zx=\frac y{xyz}+zx=zx+\frac1{zx}\geq 2$
Equality holds if and only if $zx=1$.

Substituting $xz=1$, we have the condition that $y(x+y+\frac{1}{x}) = 1$. As such, $(x+y)(y+ \frac{1}{x}) = xy + \frac{y}{x} + y^2 + 1 = 1 + 1 = 2 $.
