jmc

We have the factorization $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz).$$Let $A=x+y+z$ and $B=x^2+y^2+z^2.$ Squaring $x+y+z=A,$ we get $$x^2+y^2+z^2+2(xy+xz+yz)=A^2,$$so $xy+xz+yz=\frac{A^2-B}{2}.$ Hence, $$A\left(B-\frac{A^2-B}{2}\right)=1,$$which simplifies to $A^3+2=3AB.$

Now, by the Trivial Inequality, $$(x-y{)}^2+(x-z{)}^2+(y-z{)}^2\ge 0,$$which simplifies to $x^2+y^2+z^2\ge xy+xz+yz.$ Since $$(x+y+z)(x^2+y^2+z^2-xy-xz-yz)=1,$$we must have $A=x+y+z>0.$

From $A^3+2=3AB,$ $$B=\frac{A^3+2}{3A}.$$By AM-GM, $$\frac{A^3+2}{3A}=\frac{A^3+1+1}{3A}\ge \frac{3\sqrt[3]{A^3}}{3A}=1,$$so $B\ge 1.$

Equality occurs when $x=1,$ $y=0,$ and $z=0,$ so the minimum value is $\boxed{1}.$