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!!!


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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, Morgan Rodgers, TastyRomeo, Alex Mathers, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.


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.


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 $n<2n-1$.

  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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

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 $$\dfrac{a_{1}}{a_{2}}+\dfrac{a_{2}}{a_{3}}+\cdots+\dfrac{a_{n-1}}{a_{n}}+\dfrac{a_{n}}{a_{1}}<n-1+p+n-p=2n-1$$

  • $\begingroup$ 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

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