Prove that $\frac{x_1}{x_2}+\frac{x_2}{x_3}+\frac{x_3}{x_4}+\cdots +\frac{x_{n-1}}{x_n}+\frac{x_n}{x_1}<2n-1$ 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!!!
 A: 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.
A: 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 $n<2n-1$. 
A: 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
$$\dfrac{a_{1}}{a_{2}}+\dfrac{a_{2}}{a_{3}}+\cdots+\dfrac{a_{n-1}}{a_{n}}+\dfrac{a_{n}}{a_{1}}<n-1+p+n-p=2n-1$$
