jmc

By AM-GM, $$\begin{array}{rl}x+2y& \ge 2\sqrt{2xy},\\ y+2z& \ge 2\sqrt{2yz},\\ xz+1& \ge 2\sqrt{xz},\end{array}$$so $$(x+2y)(y+2z)(xz+1)\ge (2\sqrt{2xy})(2\sqrt{2yz})(2\sqrt{xz})=16xyz=16.$$Equality occurs when $x=2y,$ $y=2z,$ and $xz=1.$ We can solve to get $x=2,$ $y=1,$ and $z=\frac{1}{2},$ so the minimum value is $\boxed{16}.$