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(C) Given $ab^2c^3 = 64$. We want to minimize $S = \frac{1}{a} + \frac{2}{b} + \frac{3}{c}$.By the Arithmetic Mean-Geometric Mean Inequality ($AM$-$GM

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$3$

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(C) Given $ab^2c^3 = 64$. We want to minimize $S = \frac{1}{a} + \frac{2}{b} + \frac{3}{c}$.By the Arithmetic Mean-Geometric Mean Inequality ($AM$-$GM$),for positive real numbers $x_1, x_2, x_3$:$\frac{x_1 + x_2 + x_3}{3} \geq \sqrt[3]{x_1 x_2 x_3}$.Let $x_1 = \frac{1}{a}$,$x_2 = \frac{2}{b}$,and $x_3 = \frac{3}{c}$.Then $S = x_1 + x_2 + x_3 \geq 3 \sqrt[3]{x_1 x_2 x_3} = 3 \sqrt[3]{\frac{1}{a} \cdot \frac{2}{b} \cdot \frac{3}{c}} = 3 \sqrt[3]{\frac{6}{abc}}$.However,we are given $ab^2c^3 = 64$. To use $AM$-$GM$ effectively,we split the terms to match the exponents:$S = \frac{1}{a} + \frac{1}{b} + \frac{1}{b} + \frac{1}{c} + \frac{1}{c} + \frac{1}{c}$.Applying $AM$-$GM$ to these $6$ terms:$\frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{b} + \frac{1}{c} + \frac{1}{c} + \frac{1}{c}}{6} \geq \sqrt[6]{\frac{1}{a} \cdot \frac{1}{b^2} \cdot \frac{1}{c^3}} = \sqrt[6]{\frac{1}{ab^2c^3}}$.Since $ab^2c^3 = 64 = 2^6$,we have $\sqrt[6]{\frac{1}{64}} = \frac{1}{2}$.Therefore,$\frac{S}{6} \geq \frac{1}{2}$,which implies $S \geq 3$.

If $a, b, c$ are positive real numbers such that $ab^2c^3 = 64$,what is the minimum value of $(1/a + 2/b + 3/c)$?

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