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!!!

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.
