We start by setting the base case $x_1\ge x_2\ge \cdots\ge x_n\ge x_1\ge\cdots$, this is, $x_1=x_2=\cdots=x_n$. In this case it is easy to check that the equality occurs. Now we study what happens whenever we change a $\ge$ by a $<$ (changing $<$ by $\ge$ is just like doing nothing). With $x_i\geq x_{i+1}$ we have the terms $\textrm{min}(x_{i-1},x_i)+\textrm{max}(x_i, x_{i+1})$, while in the case $x_i< x_{i+1}$ we have the terms $\frac{\textrm{min}(x_{i-1},x_i)\cdot x_{i+1}}{x_i}+\frac{x_{i}\cdot\textrm{max}(x_i, x_{i+1})}{x_{i+1}}$, we now want to prove that these terms are less than the previous ones, and therefore the whole expression decreases, so the inequality is maintained. To do this we call $a=\textrm{min}(x_{i-1},x_i)$ and $b=\textrm{max}(x_i,x_{i+1})$, and let $t=\frac {x_i}{x_{i+1}}$. We rewrite those terms as $a+b$ and $at+b\frac1t$, so it is enough to prove that for any $0<a\le x_i< x_{i+1}\le b\implies \frac ab\le t< 1$, $at+b\frac 1t\le a+b$, but this follows from the fact that $f(t)=at+b\frac1t$ is convex because its second derivative is $\frac{2b}{t^3}>0$ $\forall t>0$ and $f(t)=a+b\iff t=\frac{a}{b}$ or $1$, and therefore $f(t)=at+b\frac 1t$ is below $a+b$ within the interval $t\in\left[\frac ab,1\right]$. So $\forall t\in\left[\frac ab,1\right):\ at+b\frac1t\le a+b\ \square$.