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.
Best Answer
Let $x_{k+1}=x_1$ and $x_0=x_k.$
Thus, by C-S $$\sum_{i=1}^k\frac{x_i}{x_{i-1}+x_{i+1}}=\sum_{i=1}^k\frac{x_i^2}{x_{i-1}x_i+x_{i+1}x_i}\geq\frac{\left(\sum\limits_{l=1}^kx_i\right)^2}{\sum\limits_{i=1}^k(x_{i-1}x_i+x_ix_{i+1})}=\frac{\left(\sum\limits_{l=1}^kx_i\right)^2}{2\sum\limits_{i=1}^kx_ix_{i+1}}.$$ Thus, it's enough to prove that $$\left(\sum\limits_{l=1}^kx_i\right)^2\geq4\sum\limits_{i=1}^kx_ix_{i+1}.$$
Now, for any even $k$ by AM-GM $$4\sum_{i=1}^kx_ix_{i+1}\leq4(x_1+x_3+...+x_{k-1})(x_2+x_4+...+x_k)\leq$$ $$\leq(x_1+x_3+...+x_{k-1}+x_2+x_4+...+x_k)^2=\left(\sum_{i=1}^kx_i\right)^2.$$ For odd $k$ since our inequality is cyclic, we can assume that $x_1=\min\{x_i\}$ and by AM-GM again: $$4\sum_{i=1}^kx_ix_{i+1}\leq4(x_1+x_3+...+x_{k})(x_2+x_4+...+x_{k-1})\leq$$ $$\leq(x_1+x_3+...+x_{k}+x_2+x_4+...+x_{k-1})^2=\left(\sum_{i=1}^kx_i\right)^2.$$