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 $$\frac{x+y-z}{2}+\frac{x+y-z}{2}+2z+\frac{1}{2xyz}\ge 4\sqrt[4]{\frac{(x+y-z{)}^{2}z}{4xyz}}=4$$ Hence $$x+y+z+\frac{1}{2xyz}\ge 4$$ The equality holds if $x=2$, $y=z=\frac{1}{2}$, or any its cyclic permutation.