Saudi Arabian Mathematical Competitions

Without loss of generality, assume that $z = \min \{x, y, z\}$. From the condition $x^{2} + y^{2} + z^{2} = 2(xy + yz + zx)$, we get
$$(x + y)^{2} - 2z(x + y) + z^{2} = 4xy$$
or
$$(x + y - z)^{2} = 4xy$$
Using the AM-GM inequality, we have
$$\frac{x + y - z}{2} + \frac{x + y - z}{2} + 2z + \frac{1}{2xyz} \geq 4 \sqrt[4]{\frac{(x + y - z)^{2} z}{4xyz}} = 4$$
Hence
$$x + y + z + \frac{1}{2xyz} \geq 4$$
The equality holds if $x = 2$, $y = z = \frac{1}{2}$, or any its cyclic permutation.