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(B) By the Arithmetic Mean-Geometric Mean $(AM-GM)$ inequality,for positive numbers $a, b, c/2, c/2$:$\frac{a + b + c/2 + c/2}{4} \geq \sqrt[4]{a \cdo

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$a = b = 1/4, c = 1/2$

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(B) By the Arithmetic Mean-Geometric Mean $(AM-GM)$ inequality,for positive numbers $a, b, c/2, c/2$:$\frac{a + b + c/2 + c/2}{4} \geq \sqrt[4]{a \cdot b \cdot \frac{c}{2} \cdot \frac{c}{2}}$$\frac{a + b + c}{4} \geq \sqrt[4]{\frac{abc^2}{4}}$Raising both sides to the power of $4$:$\frac{(a + b + c)^4}{256} \geq \frac{abc^2}{4}$$abc^2 \leq \frac{(a + b + c)^4}{64}$Given the maximum value of $abc^2$ is $1/64$,we have $\frac{(a + b + c)^4}{64} = \frac{1}{64}$,which implies $a + b + c = 1$.The equality holds when $a = b = c/2$.Substituting $a = b = c/2$ into $a + b + c = 1$:$c/2 + c/2 + c = 1 \implies 2c = 1 \implies c = 1/2$.Then $a = b = 1/4$.

If $a, b, c$ are three positive numbers such that the maximum value of $abc^2$ is $1/64$,then:

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