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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 $x_1, x_2, \dots, x_n$?

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7 Answers 7

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

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    $\begingroup$ So straightforward with the right tool. $\endgroup$
    – marshall
    May 27, 2013 at 16:07
  • $\begingroup$ See John's answer. $\endgroup$ May 27, 2013 at 21:00
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HINT: Use AM-GM inequality for real positive $a_i$s (where $1\le i\le n$)

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

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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.

Spoiler:

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

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  • $\begingroup$ This is just the AM-GM inequality in disguise, unless you know another way to show that $$\frac{x_i}{x_{n-i}}+\frac{x_{n-i}}{x_i}\geq 2$$ $\endgroup$ May 27, 2013 at 21:05
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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 $$x_1\cdot \frac1{x_n}+x_2\cdot\frac1{x_{n-1}}+\dots+x_n\cdot\frac1{x_1} \ge x_{i_1}\cdot \frac1{x_{i_1}}+x_{i_2}\cdot\frac1{x_{i_2}}+\dots+x_{i_n}\cdot\frac1{x_{i_n}}=n.$$

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(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.

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    $\begingroup$ I must have missed yours before I posted mine. $\endgroup$
    – John
    May 27, 2013 at 16:24
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$$\dfrac{x_1}{x_n} + \dfrac{x_2}{x_{n-1}} + \dfrac{x_3}{x_{n-2}} + \dots + \dfrac{x_n}{x_1}- n =\left(\dfrac{\sqrt{x_1}}{\sqrt{x_n}}-\dfrac{\sqrt{x_n}}{\sqrt{x_1}} \right)^2+\left(\dfrac{\sqrt{x_2}}{\sqrt{x_{n-1}}}-\dfrac{\sqrt{x_{n-1}}}{\sqrt{x_2}} \right)^2+...$$

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

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