Given, $\alpha 1, \alpha 2,\ldots,\alpha n$ are the positive numbers.
So, $(\frac{\alpha 1}{\alpha 2}) , ( \frac{\alpha 2}{\alpha 3})$ etc will be also positive numbers.
Hence applying A.M. $\geq$ G.M,
$\dfrac{\left((\frac{\alpha 1}{\alpha 2}) + (\frac {\alpha 2}{\alpha 3}) +\ldots + (\frac {\alpha n}{\alpha 1})\right)}{n}$
$\quad \quad\geq \left((\frac{\alpha 1}{\alpha 2}).(\frac{\alpha 2}{\alpha 3})\ldots(\frac{\alpha n}{\alpha 1})\right)^{\frac{1}{n}}$
$\large\Rightarrow$ $\dfrac{\left((\frac{\alpha 1}{\alpha 2}) + (\frac {\alpha 2}{\alpha 3}) +\ldots + (\frac {\alpha n}{\alpha 1})\right)}{n}$
$\quad \quad\geq n. \left((\frac{\alpha 1}{\alpha 2}).(\frac{\alpha 2}{\alpha 3})\ldots(\frac{\alpha n}{\alpha 1})\right)^{\frac{1}{n}}$
$\large\Rightarrow$ $\dfrac{\left((\frac{\alpha 1}{\alpha 2}) + (\frac {\alpha 2}{\alpha 3}) +\ldots + (\frac {\alpha n}{\alpha 1})\right)}{n}$
$\quad \quad\geq n.{1^{\frac{1}{n}}}$
$\large\Rightarrow$ $\dfrac{\left((\frac{\alpha 1}{\alpha 2}) + (\frac {\alpha 2}{\alpha 3}) +\ldots + (\frac {\alpha n}{\alpha 1})\right)}{n}$
$\quad \quad \geq n$
Hence A) is the correct answer..