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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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, 1, $\mathrm{n}$, ${ }^n \mathrm{C}_2$, 2

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Hint: $\frac{\frac{a_1}{a_2}+\frac{a_2}{a_3}+\ldots . .+\frac{a_n}{a_1}}{n} \geq \sqrt[n]{\frac{a_1}{a_2} \times \frac{a_2}{a_3} \times \ldots . \frac{a_n}{a_1}}$ $A M \geq G M$ (property) $\therefore$ Minimum value $=\mathrm{n}$ It is possible when $a_1=a_2=a_3=\ldots .=a_n$

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