Fractional Product Minimization
Let $ a, $ $ b, $ $ c, $ $ d $ be positive real numbers. Find the minimum value of
\[(1 + a)(1 + b)(1 + c)(1 + d) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right).\]
- 1
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- $\frac{a}{b}$
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- 0
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- $a^n$
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- $a_n$
- $\sqrt{}$
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Solution
By AM-GM,
\[1 + a = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} + a \ge 4 \sqrt[4]{\frac{1}{3^3} \cdot a} = 4 \sqrt[4]{\frac{a}{27}}.\]Similarly,
\begin{align*}
1 + b &\ge 4 \sqrt[4]{\frac{b}{27}}, \\
1 + c &\ge 4 \sqrt[4]{\frac{c}{27}}, \\
1 + d &\ge 4 \sqrt[4]{\frac{d}{27}}.\end{align*}Also by AM-GM,
\[\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \ge 4 \sqrt[4]{\frac{1}{abcd}}.\]Multiplying all these inequalities, we get
\begin{align*}
(1 + a)(1 + b)(1 + c)(1 + d) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right) &\ge 4 \sqrt[4]{\frac{a}{27}} \cdot 4 \sqrt[4]{\frac{b}{27}} \cdot 4 \sqrt[4]{\frac{c}{27}} \cdot 4 \sqrt[4]{\frac{d}{27}} \cdot 4 \sqrt[4]{\frac{1}{abcd}} \\
&= \frac{1024}{27}.\end{align*}Equality occurs when $ a = b = c = d = \frac{1}{3}, $ so the minimum value is $ \boxed{\frac{1024}{27}} $.
Freeflows