# Let $x,y,z>0; ~ xyz(x+y+z)=1$. Show that $(x+y)(y+z)\ge 2$

Assuming positive real values of $x,y,z$, and that $xyz(x+y+z)=1$, how can we prove that $(x+y)(y+z)\ge 2$

I tried using the AM-GM inequality but as if I were looping. I'm not sure if C-S or other inequality methods could help. Thanks for helping me.

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$$(x+y)(y+z)=xy+y^2+zx+yz=y(x+y+z)+zx=\frac y{xyz}+zx=zx+\frac1{zx}\ge 2\sqrt{\frac1{zx}zx}=2$$ using A.M.$\ge$ G.M. for $z,x>0$
I reached until $zx +\frac{1}{zx}$ but never thought of completing your answer. Thanks! – Tariq Jan 9 '13 at 11:25