solution: By AM-GM,
  \[a + b \ge 2 \sqrt{ab},\]so $(a + b)^2 \ge 4ab.$
  
  Also by AM-GM,
  \[(a + 2c) + (b + 2c) \ge 2 \sqrt{(a + 2c)(b + 2c)},\]so $(a + b + 4c)^2 \ge 4(a + 2c)(b + 2c).$
  
  Hence,
  \begin{align*}
  (a + b)^2 + (a + b + 4c)^2 &\ge 4ab + 4(a + 2c)(b + 2c) \\
  &= 8ab + 8ac + 8bc + 16c^2 \\
  &= 8(ab + ac + bc + 2c^2).
  \end{align*}By AM-GM,
  \begin{align*}
  ab + ac + bc + 2c^2 &= \frac{ab}{2} + \frac{ab}{2} + ac + bc + 2c^2 \\
  &\ge 5 \sqrt[5]{\frac{ab}{2} \cdot \frac{ab}{2} \cdot ac \cdot bc \cdot 2c^2} \\
  &= 5 \sqrt[5]{\frac{a^3 b^3 c^4}{2}}.
  \end{align*}Also by AM-GM,
  \begin{align*}
  a + b + c &= \frac{a}{2} + \frac{a}{2} + \frac{b}{2} + \frac{b}{2} + c \\
  &\ge 5 \sqrt[5]{\frac{a}{2} \cdot \frac{a}{2} \cdot \frac{b}{2} \cdot \frac{b}{2} \cdot c} \\
  &= 5 \sqrt[5]{\frac{a^2 b^2 c}{16}}.
  \end{align*}Hence,
  \begin{align*}
  \frac{(a + b + c)[(a + b)^2 + (a + b + 4c)^2]}{abc} &\ge 8 \cdot \frac{5 \sqrt[5]{\frac{a^2 b^2 c}{16}} \cdot 5 \sqrt[5]{\frac{a^3 b^3 c^4}{2}}}{abc} \\
  &= 100.
  \end{align*}Equality occurs when $a = b = 2$ and $c = 1,$ so the minimum value is $\boxed{100}.$
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  Let $a,$ $b,$ and $c$ be positive real numbers.  Find the minimum value of
  \[\frac{(a + b + c)[(a + b)^2 + (a + b + 4c)^2]}{abc}.\]
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: 100
extra_info.question: Let $a,$ $b,$ and $c$ be positive real numbers.  Find the minimum value of
  \[\frac{(a + b + c)[(a + b)^2 + (a + b + 4c)^2]}{abc}.\]
extra_info.split: test