Since $\sum\limits_{k=1}^{n-1}min(x_k,x_{k+1})\le \sum\limits_{k=1}^{n-1}x_k$ and $\sum\limits_{k=1}^{n-1}min(x_k,x_{k+1})\le \sum\limits_{k=1}^{n-1}x_{k+1}$ So according condition，$min(x_1,x_n)=\sum\limits_{k=1}^{n-1}min(x_k,x_{k+1})\le \sum\limits_{k=2}^{n-1}x_k+min(x_1,x_n)$ Therefore, $\sum_{k=2}^{n-1} x_k \ge 0$

I think this solution is not correct

We know that $min(a,b)=\frac{a+b-\begin{vmatrix}b-a\end{vmatrix}}{2}$, and denote: S=$\sum_{k=2}^{n-1}x_{k},$so we have $\sum_{k=1}^{n-1}min(x_{k},x_{k+1})$=$\frac{1}{2}\left(\sum_{k=1}^{n-1}(x_{k}+x_{k+1})\right)-\frac{1}{2}\sum_{k=1}^{n-1}\begin{vmatrix}x_{k+1}-x_{k}\end{vmatrix}=S+\frac{x_{n}+x_{1}}{2}$-$\frac{1}{2}$$\sum_{k=1}^{n}\begin{vmatrix}x_{k+1}-x_{k}\end{vmatrix}$=$min(x_{1},x_{n})=$ $\frac{x_{n}+x_{1}-\begin{vmatrix}x_{n}-x_{1}\end{vmatrix}}{2}$so we got $2S=\sum_{k=1}^{n-1}\begin{vmatrix}x_{k+1}-x_{k}\end{vmatrix}-\begin{vmatrix}x_{n}-x_{1}\end{vmatrix}\geq0,$because $\begin{vmatrix}a+b\end{vmatrix}\leq\begin{vmatrix}a\end{vmatrix}+\begin{vmatrix}b\end{vmatrix}$ for $n-1$ terms too!.