Inequality $\frac{x_1}{x_k+x_2}+\frac{x_2}{x_1+x_3}+\dots+\frac{x_k}{x_{k-1}+x_1}\geq 2.$

Question

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.

Answer 1

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: $$4\sum_{i=1}^kx_ix_{i+1}\leq4(x_1+x_3+...+x_{k})(x_2+x_4+...+x_{k-1})\leq$$ $$\leq(x_1+x_3+...+x_{k}+x_2+x_4+...+x_{k-1})^2=\left(\sum_{i=1}^kx_i\right)^2.$$

Answer 2

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.