mavropnevma wrote: For a given integer $n\geq 3$, determine the range of values for the expression \[ E_n(x_1,x_2,\ldots,x_n) := \dfrac {x_1} {x_2} + \dfrac {x_2} {x_3} + \cdots + \dfrac {x_{n-1}} {x_n} + \dfrac {x_n} {x_1}\] over real numbers $x_1,x_2,\ldots,x_n \geq 1$ satisfying $|x_k - x_{k+1}| \leq 1$ for all $1\leq k \leq n-1$. Do also determine when the extremal values are achieved. (Dan Schwarz) It is clear that $\min E_n=n$ when $x_1=x_2=\cdots =x_n.$ Now, we will prove that \[\frac{x_{n-1}}{x_n}+\frac{x_n}{x_1} \le \frac{x_{n-1}}{x_1} +\frac{2n-1}{n},\quad \forall n \ge 2.\quad (1)\] This inequality can be written as \[\frac{(x_n-x_1)(x_n-x_{n-1})}{x_1x_n} \le \frac{n-1}{n}.\quad (2)\] If $\min \{x_1,\, x_{n-1}\} \le x_n \le \max \{x_1,\, x_{n-1}\},$ then the above inequality is obviously true. In the case $x_n \ge \max\{x_1,\, x_{n-1}\},$ we have $|x_n-x_{n-1}| \le 1,$ so $x_n \le x_{n-1}+1.$ On the other hand, we also have \[x_n=(x_n-x_{n-1})+\cdots +(x_2-x_1)+x_1 \le x_1+n-1.\] It follows that \[\mathrm{LHS}_{(2)} \le \frac{x_n-x_1}{x_nx_1} \le \frac{n-1}{x_1(x_1+n-1)} \le \frac{n-1}{n}=\mathrm{RHS}_{(2)}.\] Let us consider now the case $x_n \le \min\{x_1,\,x_{n-1}\}.$ Since $|x_{n}-x_{n-1}| \le 1,$ we have $x_{n-1} \le x_n+1.$ Beside, we also have \[x_1=(x_1-x_2)+\cdots +(x_{n-1}-x_n)+x_ n \le x_n+n-1.\] Therefore, \[\begin{aligned} \mathrm{LHS}_{(2)}&=\frac{(x_1-x_n)(x_{n-1}-x_n)}{x_1x_n} \le \frac{x_1-x_n}{x_1x_n} \\ &\le \frac{n-1}{x_n(x_n+n-1)} \le \frac{n-1}{n}=\mathrm{RHS}_{(2)}.\end{aligned}\] So $(1)$ is proved. Now, we will use induction and $(1)$ to prove that \[E_n \le \frac{1}{2}+\cdots +\frac{n-1}{n}+\frac{n}{1}\] for any $n \ge 2.$ For $n=2,$ the inequality becomes \[\frac{x_1}{x_2}+\frac{x_2}{x_1} \le \frac{5}{2},\] or \[\frac{(2x_1-x_2)(2x_2-x_1)}{x_1x_2} \ge 0,\] which is true because $2x_1-x_2 \ge 1+x_1-x_2 \ge 0$and $2x_2-x_1 \ge 1+x_2-x_1 \ge 0.$ Assume that the inequality holds for $n-1$ $(n \ge 3),$ i.e. \[\frac{x_1}{x_2}+\cdots +\frac{x_{n-1}}{x_1} \le \frac{1}{2}+\cdots +\frac{n-2}{n-1}+\frac{n-1}{1}.\] Using the induction hypothesis and $(1),$ we have \[\begin{aligned}E_n&=E_{n-1}+\frac{x_{n-1}}{x_n}+\frac{x_n}{x_1}-\frac{x_{n-1}}{x_1}\\ & \le \left( \frac{1}{2}+\cdots +\frac{n-2}{n-1}+\frac{n-1}{1}\right)+\frac{2n-1}{n}\\ &=\frac{1}{2}+\cdots +\frac{n-1}{n}+\frac{n}{1}.\end{aligned}\] So, we have proved that \[E_n \le \frac{1}{2}+\cdots +\frac{n-1}{n}+\frac{n}{1}.\] I will leave the equality case here, I think any readers can work on this by their own.

Isn't this related ? http://www.artofproblemsolving.com/Forum/viewtopic.php?p=341879&sid=dd67c97e0cd6d7b16a702626758a28f5#p341879