solution: 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 $\boxed{16}.$
prompt.content: <|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
  
prompt.role: user
extra_info.answer: 16
extra_info.question: Let $x,$ $y,$ and $z$ be positive real numbers such that $xyz = 2.$  Find the minimum value of
  \[x^4 + 4y^2 + 4z^4.\]
extra_info.split: test