By AM-GM,
\begin{align*}
x^4 + 4y^2 + 4z^4 &= x^4 + 2y^2 + 2y^2 + 4z^4 \\
&\ge 4 \sqrt[4]{(x^4)(2y^2)(2y^2)(4z^4)} \\
&= 8xyz \\
&= 16.
\end{align*}Equality occurs when $x^4 = 2y^2 = 4z^2.$  Using the condition $xyz = 2,$ we can solve to get $x = y = \sqrt{2}$ and $z = 1,$ so the minimum value is $\boxed{16}.$
<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Let $x,$ $y,$ and $z$ be positive real numbers such that $xyz = 2.$  Find the minimum value of
\[x^4 + 4y^2 + 4z^4.\]
Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
<|im_start|>assistant

user
16
Let $x,$ $y,$ and $z$ be positive real numbers such that $xyz = 2.$  Find the minimum value of
\[x^4 + 4y^2 + 4z^4.\]
test