$\rm xy+yz+zx\geq 3\Leftrightarrow (xy+yz+zx)(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z})\geq 3(x+y+z)\Leftrightarrow \dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\geq x+y+z \Leftrightarrow $ $\rm \Leftrightarrow \dfrac{(xy)^2+(yz)^2+(zx)^2}{xyz}\geq x+y+z\Leftrightarrow (xy)^2+(yz)^2+(zx)^2\geq (x+y+z)xyz$, which is true.

parmenides51 wrote: Let $x, y$ and $z$ be positive real numbers such that $x + y + z = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ . Prove that $xy + yz + zx \ge 3$. By AM-GM, $$xy + yz + zx=\frac{(xy + yz + zx)^2}{xy + yz + zx} \ge \frac{xyz(x+ y+ z)}{xy + yz + zx} =3$$