Q) If $a_{1},a_{2},a_{3},....a_{n}$ are positive real numbers,then

$\frac{a_{1}}{a_{2}} + \frac{a_{2}}{a_{3}}+......+\frac{a_{n-1}}{a_{n}}+\frac{a_{n}}{a_{1}}$

is always

(A) $\geq n \;$ (B) $\leq n\;$ (C) $\leq n^{1/n}\;$ (D) none of the above

I have taken some small values of $n$ like $n=2,3$ and sample inputs as $(a_1,a_2,a_3)=(2,3,5)$,$(a_1,a_2,a_3)=(1/2,1/3,1/5)$ , $(a_1,a_2,a_3)=(0.1,0.2,0.3)$, $(a_1,a_2)=(2,3)$ for which I am getting answer as $(A)$ but I am not sure that it is always true.If it is not true always then answer will be $(D).$ I don't know how to prove it formally. Please help.

  • 2
    $\begingroup$ Use AM-GM. First will be right answer. $\endgroup$ Mar 17, 2019 at 7:09
  • $\begingroup$ Thank you @Love Invariants, got it..Now I can continue. $\endgroup$
    – ankit
    Mar 17, 2019 at 7:13
  • $\begingroup$ @LoveInvariants Why are you answering in a comment? $\endgroup$
    – Arthur
    Mar 17, 2019 at 7:15
  • $\begingroup$ @Arthur-Because a hint would satisfy. OP should do the rest of the work. It helps askers very much. $\endgroup$ Mar 17, 2019 at 7:16
  • $\begingroup$ @LoveInvariants Hints are still answers more than they are clarification requests. So they belong in answer posts, not in comments. Also, you have a direct answer to the question by saying which alternative was right, so calling it just a hint isn't quite right. $\endgroup$
    – Arthur
    Mar 17, 2019 at 7:18

2 Answers 2


Hint: Use AM-GM. First will be right answer.
AM-GM inequality


This may be solved using the rearrangement inequality. We have the two sequences $$ a_1, a_2, a_3,\ldots,a_n\\ \frac1{a_1}, \frac1{a_2},\frac1{a_3}, \ldots,\frac1{a_n} $$ If we multiply each $a_i$ with $\frac1{a_i}$, then the sum is $n$. By the rearrangement inequality, this is the smallest possible value we can get, so rearranging must give us a result $\geq n$.

Or one may use the AM-GM inequality to get the same result.


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