Does anybody have any idea as how to proceed?

if $n=3$, we get $min\ge max$

This actually does not occur due to the n+2 in the denominator.

mssmath wrote: This actually does not occur due to the n+2 in the denominator. he edited, but the main problem is that $ n\rightarrow \infty \Rightarrow \text{min}\ge2\ \text{max}$

To respond to this maybe the problem is with the max an min reversed.

I've edited the problem . The inequality sign was flipped.Sorry about that.

$ \min\left ( x_1,\frac{1}{x_1}+x_2, \cdots,\frac{1}{x_{n-1}}+x_n,\frac{1}{x_n} \right )\leq 2\cos \frac{\pi}{n+2} \le \max\left ( x_1,\frac{1}{x_1}+x_2, \cdots,\frac{1}{x_{n-1}}+x_n,\frac{1}{x_n} \right ). $ this is the correct statement,checked with the RMC book. Now,let's solve the correct version. We assume that $\min\left ( x_1,\frac{1}{x_1}+x_2, \cdots,\frac{1}{x_{n-1}}+x_n,\frac{1}{x_n} \right )> 2\cos a$ ,where $a=\frac{\pi}{n+2}.$ Now,we can inductively prove that $x_i>\frac{\sin (i+1)a}{\sin ia}$;the case $i=1$ is immediate,while the inequality for $i+1$ can be derived from the one for $i$ by means of $x_{i+1}+\frac{1}{x_i}>2\cos a$ by standard trig manipulations. However,for $i=n$ this implies that $x_n>\frac{\sin (n+1)a}{\sin na}=\frac{\sin a}{\sin 2a}=\frac{1}{2\cos a}$,which contradicts $\frac{1}{x_n}>2\cos a$,thus ending the proof. The other inequality is treated in the same way.