Inequality $\frac{x_1}{x_k+x_2}+\frac{x_2}{x_1+x_3}+\dots+\frac{x_k}{x_{k-1}+x_1}\geq 2.$ Prove that for any positive numbers $x_1, x_2,\dots, x_k$ $(k>3)$ the following inequality is true: $$\frac{x_1}{x_k+x_2}+\frac{x_2}{x_1+x_3}+\dots+\frac{x_k}{x_{k-1}+x_1}\geq 2.$$
The case when $k=4$ can be reduced to the inequality $a+\frac{1}{a}\geq 2$ so in this case everything is fine.
But for $k>4$ I do not see the way how to prove that inequality.
Can you show the solution please?
P.S. It seems like a Nesbitt inequality which states that if $a,b,c>0$ then $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq \frac{3}{2}$. I do know how to prove that one. We can make substitution $b+c=x, a+c=y, a+b=z$ and the rest is quite obvious.

Incrementing $k$ from $j$ to $j+1$ increases the left-hand side by$$\frac{x_{j+1}}{x_j+x_1}+\color{red}{\frac{x_1}{x_{j+1}+x_2}-\frac{x_1}{x_j+x_2}}+\color{blue}{\frac{x_j}{x_{j-1}+x_{j+1}}-\frac{x_j}{x_{j-1}+x_1}}.$$With the above coloration, each monochromatic part is non-negative if we label the $j+1$ variables so $x_{j+1}$ is minimal. While we can't arbitrarily permute the $x_i$, we can, through rotation of the list of fractions, place the one with least numerator last, so this $x_{j+1}$-is-least convention handles the general case of the inductive step.
