ISI2004-MIII: 11

Question

If $\alpha 1,\alpha 2,\dots,\alpha n$ are the positive numbers then

$\frac{a1}{a2}+\frac{a2}{a3}+\dots+\frac{an-1}{an}+\frac{an}{a1}$ is always

• A. $\geq n$
• B. $\leq n$
• C. $\leq n^{\frac{1}{2}}$
• D. None of the above

Comments

$\frac{1}{2}$+$\frac{2}{3}$+$\frac{3}{1}$ =$\frac{25}{6}$ $\geq$ 3

so Ans:A

Answer 1

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..

Comments

@habib,we can also check it byt putting value for n..right??

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Yes we can do that @Akriti sood...

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thankyou..!