8
$\begingroup$

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!!!

3 Answers 3

5

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.

4

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
    But $x_n/x_1$ is not of the form $x_i/x_{i+1}$ – 2017-01-10
  • 0
    edited. good now? – 2017-01-10
  • 0
    Why can you assume that $x_n/x_1 < 1$? – 2017-01-10
  • 0
    I can assume anything I want, as long as I also cover the case where that particular assumption does not hold. – 2017-01-10
  • 1
    @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. – 2017-01-10
  • 1
    @MartinR, actually you are right, it is not given that $|x_1-x_n|<1$. will have to modify it later. – 2017-01-10
  • 0
    @Winther: Of course *some* fraction must be less than one. But if (for example) $x_3/x_4 < 1$ is assumed then above method can be used to estimate all remaining fractions *except* $x_n/x_1$. – 2017-01-10
  • 1
    @MartinR Ah, I was a bit blind there, thanks for clearing that up. Your point is absolutely valid then. – 2017-01-10
2

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}

(2):define set $A=\{k|a_{k}>a_{k+1}\}$,Assmue that $|A|=p$,for $k\in A$,we have $$a_{k+1}

  • 0
    You could also treat case (1) as a special case of case (2) with $A$ being empty and $p=0$. – 2017-01-11
  • 0
    @MartinR,Thanks – 2017-01-11