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(B) By the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality, for positive real numbers $x_1, x_2, \ldots, x_n$:$\frac{x_1 + x_2 + \ldots + x_n

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$n$

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(B) By the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality, for positive real numbers $x_1, x_2, \ldots, x_n$:$\frac{x_1 + x_2 + \ldots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \ldots x_n}$Let $x_1 = \frac{a_1}{a_2}, x_2 = \frac{a_2}{a_3}, \ldots, x_n = \frac{a_n}{a_1}$.Then, the product $x_1 x_2 \ldots x_n = \frac{a_1}{a_2} 	imes \frac{a_2}{a_3} 	imes \ldots 	imes \frac{a_n}{a_1} = 1$.Substituting this into the inequality:$\frac{\frac{a_1}{a_2} + \frac{a_2}{a_3} + \ldots + \frac{a_n}{a_1}}{n} \geq \sqrt[n]{1} = 1$Therefore, $\frac{a_1}{a_2} + \frac{a_2}{a_3} + \ldots + \frac{a_n}{a_1} \geq n$.The minimum value is $n$, which occurs when $a_1 = a_2 = \ldots = a_n$.

Let $a_1, a_2, a_3, \ldots, a_n$ be positive real numbers. Then the minimum value of $\frac{a_1}{a_2}+\frac{a_2}{a_3}+\ldots+\frac{a_n}{a_1}$ is

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