Minimal value of the product $(x+y)(y+z)$ 
Find the min. value of the product $\left(x + y\right)\left(y + z\right)$ given that $xyz\left(x + y + z\right) = 1$ and $x,y,z > 0$.

I want to solve this problem using geometry of a triangle:

*

*Let $a,b,c$ be the sides of the triangle and we can substitute
$a = x + y\,,\,\, b = y + z\,,\,\, c = z + x$.

*$A^{2} = xyz\left(x + y + z\right)$ is just a squared area of a triangle and $\left(x + y\right)\left(y + z\right)$ is a part of another formula of triangle  $$
\mbox{i.e.}\quad A = {1 \over 2}\,ab\sin\left(\theta\right)\quad \mbox{or}\quad
A = {1 \over 2}\left(x + y\right)\left(y + z\right)\sin\left(\theta\right)
$$
I' ve tried the same with other formulas but I don't know how to "extract" inequality.

Before the solution I would appreciate that you give a hint, and then a solution. Any help appreciated.

Let $a=x+y, \, b=y+z ,\, c = z+x, \, p = \frac{a+b+c}{2},$ then
$$(x+y)(y+z) = ab,$$
and
$$1 = \sqrt{xyz(x+y+z)} = \sqrt{p(p-a)(p-b)(p-c)} = S = \frac{1}{2} ab \sin C.$$
But $\sin C \leqslant 1,$ so
$$2 = ab \sin C \leqslant ab.$$
Therefore $ab \geqslant 2.$ Equality hold for $x = 1, \, y = \sqrt 2  -1, \, z = 1.$
P/s. Using the AM-GM inequality, we have
$$(x+y)(y+z) = xz+y(x+y+z) \geqslant 2\sqrt{xyz(x+y+z)} = 2.$$
