20. Find the least value of the expression $(x+y)(y+z)$, given that $x, y, z$ are positive real numbers satisfying the equation
$$
x y z(x+y+z)=1 \text {. }
$$
20. Answer: 2
Observe that
$$
(x+y)(y+z)=x y+x z+y^{2}+y z=y(x+y+z)+x z=\frac{1}{x z}+x z \geq 2
$$
where the equality holds if and only if $x z=1$. Let $x=z=1$ and $y=\sqrt{2}-1$, then we have the minimum value 2 for $(x+y)(y+z)$.