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Let $x_1,x_2,x_3,\cdots ,x_n (n\ge2)$ be real numbers greater than $1.$ Suppose that $|x_i-x_{i+1}|<1$ for $i=1,2,3,\cdots,(n-1)$.

Prove that $$\frac{x_1}{x_2}+\frac{x_2}{x_3}+\frac{x_3}{x_4}+\cdots +\frac{x_{n-1}}{x_n}+\frac{x_n}{x_1}<2n-1$$

Please help!!!

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closed as off-topic by Carl Mummert, Morgan Rodgers, TastyRomeo, Alex Mathers, user91500 Feb 28 '17 at 9:54

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Set $d_i = x_{i+1} - x_i$ for $i = 1, \ldots, n-1$. Then $$ \frac{x_n}{x_1} = 1 + \sum_{i=1}^{n-1} \frac{d_i}{x_1} \quad , \quad \frac{x_i}{x_{i+1}} = 1 - \frac{d_i}{x_{i+1}} $$ and therefore $$ \frac{x_n}{x_1} + \sum_{i=1}^{n-1} \frac{x_i}{x_{i+1}} = n + \sum_{i=1}^{n-1} d_i \left( \frac{1}{x_1} - \frac{1}{x_{i+1}} \right) \\ \le n + \sum_{i=1}^{n-1} \lvert d_i \lvert \left\lvert \frac{1}{x_1} - \frac{1}{x_{i+1}} \right \rvert < n + (n-1) = 2n-1 $$ because

  • $\lvert d_i \lvert < 1$ and
  • all numbers $1/x_i$ are in the interval $(0, 1]$ so that the (absolute value of) any difference $1/{x_1} - 1/{x_{i+1}}$ is less than one.

The inequality can actually be improved a bit. For fixed $i$ set $m = \min(x_1, x_{i+1})$ and $M = \max(x_1, x_{i+1})$. Then $M < m + i$ and therefore $$ \left\lvert \frac{1}{x_1} - \frac{1}{x_{i+1}} \right \rvert = \frac 1m - \frac 1M < \frac 1m - \frac{1}{m+i} = \frac{i}{m(m+i)} \le \frac{i}{i+1} $$ and the above method gives $$ \frac{x_n}{x_1} + \sum_{i=1}^{n-1} \frac{x_i}{x_{i+1}} < n + \sum_{i=1}^{n-1}\frac{i}{i+1} = (2n-1) - \sum_{i=2}^{n} \frac 1i \, . $$ Choosing $x_i = i + (i-1) \varepsilon$ shows that this bound is sharp.

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First, suppose that at least one of the fractions is strictly less than 1. Without loss of generality, we can assume that $\dfrac{x_n}{x_1}<1$. Then, $$\dfrac{x_n}{x_1}+\sum_{i=1}^{n-1}\dfrac{x_i}{x_{i+1}}<1+\sum_{i=1}^{n-1}(1+\dfrac{x_i-x_{i+1}}{x_{i+1}})<1+2(n-1) = 2n-1.$$

Now if, all of the fractions are at least 1, then they are all equal to 1 because their product is 1. In that case, the sum is $n$ and readily satisfies $n<2n-1$.

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    $\begingroup$ But $x_n/x_1$ is not of the form $x_i/x_{i+1}$ $\endgroup$ – Martin R Jan 10 '17 at 9:49
  • $\begingroup$ edited. good now? $\endgroup$ – dezdichado Jan 10 '17 at 9:51
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    $\begingroup$ @Winther: Perhaps I am overlooking something, but I think it does matter. All terms $x_i/x_{i+1}$ can be estimated by $2$ but not the term $x_n/x_1$ because $|x_1 - x_n| < 1$ is not given. $\endgroup$ – Martin R Jan 10 '17 at 10:12
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    $\begingroup$ @MartinR, actually you are right, it is not given that $|x_1-x_n|<1$. will have to modify it later. $\endgroup$ – dezdichado Jan 10 '17 at 10:25
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    $\begingroup$ @MartinR Ah, I was a bit blind there, thanks for clearing that up. Your point is absolutely valid then. $\endgroup$ – Winther Jan 10 '17 at 12:54
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Proof: I get this proof for two case

(1): if for all $k=1,2,\cdots,n-1$,have $a_{k}\le a_{k+1}$,then we have $$a_{k}\le a_{k+1}<a_{k}+1$$ so we have $$a_{i}<a_{i-1}+1<a_{i-2}+2<\cdots<a_{1}+(i-1)$$ then we have $$\dfrac{a_{1}}{a_{2}}+\dfrac{a_{2}}{a_{3}}+\cdots+\dfrac{a_{n-1}}{a_{n}}+\dfrac{a_{n}}{a_{1}}<(n-1)+\dfrac{a_{1}+(n-1)}{a_{1}}<(n-1)+1+(n-1)=2n-1$$

(2):define set $A=\{k|a_{k}>a_{k+1}\}$,Assmue that $|A|=p$,for $k\in A$,we have $$a_{k+1}<a_{k}<a_{k+1}+1,\dfrac{a_{k}}{a_{k+1}}<\dfrac{a_{k+1}+1}{a_{k+1}} <1+\dfrac{1}{a_{k+1}}<2$$ then we have $$\dfrac{a_{1}}{a_{2}}+\dfrac{a_{2}}{a_{3}}+\cdots+\dfrac{a_{n-1}}{a_{n}}<2p+(n-1-p)=n-1+p$$ and $$\dfrac{a_{n}}{a_{1}}=\dfrac{(a_{n}-a_{n-1})+(a_{n-1}-a_{n-2})+\cdots+(a_{2}-a_{1})+a_{1}}{a_{1}}<\dfrac{(n-1-p)+a_{1}}{a_{1}}<(n-1-p)+1=n-p$$ so for this case also have $$\dfrac{a_{1}}{a_{2}}+\dfrac{a_{2}}{a_{3}}+\cdots+\dfrac{a_{n-1}}{a_{n}}+\dfrac{a_{n}}{a_{1}}<n-1+p+n-p=2n-1$$

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  • $\begingroup$ You could also treat case (1) as a special case of case (2) with $A$ being empty and $p=0$. $\endgroup$ – Martin R Jan 11 '17 at 12:44
  • $\begingroup$ @MartinR,Thanks $\endgroup$ – math110 Jan 11 '17 at 12:48

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