Let $x,$ $y,$ and $z$ be positive real numbers such that $xyz = 1.$  Find the minimum value of
\[(x + 2y)(y + 2z)(xz + 1).\]
By AM-GM,
\begin{align*}
x + 2y &\ge 2 \sqrt{2xy}, \\
y + 2z &\ge 2 \sqrt{2yz}, \\
xz + 1 &\ge 2 \sqrt{xz},
\end{align*}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 $\boxed{16}.$