This is the equality case of AM-HM applied to $x,$ $2y,$ $4z,$ and 8, so they must be equal.  Hence, $(x,y,z) = (8,4,2).$
If $x,$ $y,$ and $z$ are positive real numbers such that
\[\frac{x + 2y + 4z + 8}{4} = \frac{4}{\frac{1}{x} + \frac{1}{2y} + \frac{1}{4z} + \frac{1}{8}}.\]Find the ordered triple $(x,y,z).$