Quadratic Product Minimization
Let $ x, $ $ y, $ and $ z $ be positive real numbers. Then the minimum value of
\[\frac{(x^4 + 1)(y^4 + 1)(z^4 + 1)}{xy^2 z}\]is of the form $ \frac{a \sqrt{b}}{c}, $ for some positive integers $ a, $ $ b, $ and $ c, $ where $ a $ and $ c $ are relatively prime, and $ b $ is not divisible by the square of a prime. Enter $ a + b + c $.
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Solution
By AM-GM,
\begin{align*}
\frac{x^4 + 1}{x} &= x^3 + \frac{1}{x} \\
&= x^3 + \frac{1}{3x} + \frac{1}{3x} + \frac{1}{3x} \\
&\ge 4 \sqrt[4]{x^3 \cdot \frac{1}{3x} \cdot \frac{1}{3x} \cdot \frac{1}{3x}} \\
&= \frac{4}{\sqrt[4]{27}}.\end{align*}Similarly,
\[\frac{z^4 + 1}{z} \ge \frac{4}{\sqrt[4]{27}}.\]Again by AM-GM,
\[\frac{y^4 + 1}{y^2} = y^2 + \frac{1}{y^2} \ge 2 \sqrt{y^2 \cdot \frac{1}{y^2}} = 2.\]Therefore,
\[\frac{(x^4 + 1)(y^4 + 1)(z^4 + 1)}{xy^2 z} \ge \frac{4}{\sqrt[4]{27}} \cdot 2 \cdot \frac{4}{\sqrt[4]{27}} = \frac{32 \sqrt{3}}{9}.\]Equality occurs when $ x^3 = \frac{1}{3x}, $ $ y^2 = \frac{1}{y^2}, $ and $ z^3 = \frac{1}{3z} $. We can solve, to get $ x = \frac{1}{\sqrt[4]{3}}, $ $ y = 1, $ and $ z = \frac{1}{\sqrt[4]{3}}, $ so the minimum value is $ \frac{32 \sqrt{3}}{9} $. The final answer is $ 32 + 3 + 9 = \boxed{44} $.
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