Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was recently asked this question which stumped me.

How can you show $\dfrac{x_1}{x_n} + \dfrac{x_2}{x_{n-1}} + \dfrac{x_3}{x_{n-2}} + \dots + \dfrac{x_n}{x_1} \geq n$ for any positive reals $x_1, x_2, \dots, x_n$?

share|improve this question

7 Answers 7

up vote 21 down vote accepted

By the AM-GM inequality, we have $$\frac{x_1/x_n+x_2/x_{n-1}+\cdots+x_n/x_1}{n}\geq\sqrt[n]{\frac{x_1}{x_n}\frac{x_2}{x_{n-1}}\cdots\frac{x_n}{x_1}}=\sqrt[n]{1}=1.$$

share|improve this answer
So straightforward with the right tool. –  marshall May 27 '13 at 16:07
See John's answer. –  thethuthinnang May 27 '13 at 21:00

HINT: Use AM-GM inequality for real positive $a_i$s (where $1\le i\le n$)

$$\frac{\sum_{1\le r\le n} a_i}n\ge \sqrt[n]{\prod_{ 1\le r\le n}a_i}$$

share|improve this answer

Dude. You shouldn't need all those fancy things. Just pair up each $\frac{x_i}{x_{n-i}} + \frac{x_{n-i}}{x_i}$ and examine each of those separately. Take into account even $n$ and odd $n$, since there'll be an unpaired fraction when $n$ is odd.


When $n$ is odd, identify what the unpaired fraction is, then it's value will be obvious.

share|improve this answer
This is just the AM-GM inequality in disguise, unless you know another way to show that $$\frac{x_i}{x_{n-i}}+\frac{x_{n-i}}{x_i}\geq 2$$ –  Alex Becker May 27 '13 at 21:05

(Intuitive approach with an hidden use of AM-GM)If $n$ is even you can put togheter couples which are equidistant from $n/2$ and then $x_i/x_{n+1-i}+x_{n+1-i}/x_i \geq2$ (and you have $n/2$ of such couples); if n is odd it's almost the same since you'll have a central 1.

share|improve this answer
I must have missed yours before I posted mine. –  John May 27 '13 at 16:24

HINT: Rearrangement inequality (http://en.wikipedia.org/wiki/Rearrangement_inequality).

For some permutation $(i_1,i_2,\cdots,i_n)$ of $(1,2,\cdots,n)$, the following holds: $$x_{i_1}\ge x_{i_2}\ge \dots\ge x_{i_{n-1}}\ge x_{i_n}(>0),$$ $$\frac1{x_{i_n}}\ge \frac1{x_{i_{n-1}}}\ge \dots \ge \frac1{x_{i_2}}\ge \frac1{x_{i_1}}.$$

It follows from the rearrangement inequality that $$x_1\cdot \frac1{x_n}+x_2\cdot\frac1{x_{n-1}}+\dots+x_n\cdot\frac1{x_1} \ge x_{i_1}\cdot \frac1{x_{i_1}}+x_{i_2}\cdot\frac1{x_{i_2}}+\dots+x_{i_n}\cdot\frac1{x_{i_n}}=n.$$

share|improve this answer

$$\dfrac{x_1}{x_n} + \dfrac{x_2}{x_{n-1}} + \dfrac{x_3}{x_{n-2}} + \dots + \dfrac{x_n}{x_1}- n =\left(\dfrac{\sqrt{x_1}}{\sqrt{x_n}}-\dfrac{\sqrt{x_n}}{\sqrt{x_1}} \right)^2+\left(\dfrac{\sqrt{x_2}}{\sqrt{x_{n-1}}}-\dfrac{\sqrt{x_{n-1}}}{\sqrt{x_2}} \right)^2+...$$

share|improve this answer

Hint: $\frac a b + \frac b a\ge2$

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.