2024 CGMO

Proof. The minimum value is $\frac{3}{5}n$. On one hand, taking $x_1=x_2=\cdots =x_n=1$, we have $$\sum _{k=1}^n\frac{1+x_k^2+x_k^4}{1+x_{k+1}+x_{k+1}^2+x_{k+1}^3+x_{k+1}^4}=\frac{3}{5}n.$$ On the other hand, by the AM-GM inequality, for any $1\le k\le n$, we have $x_k^2+x_k^4\ge 2x_k^3$ and $1+x_k^2\ge 2x_k$. Noting that $1+x_k^4\ge x_k+x_k^3$ (since $(x_k-1{)}^2(x_k^2+x_k+1)\ge 0$), it follows that $2(1+x_k^2+x_k^4)\ge 3(x_k+x_k^3)$. Thus, $5(1+x_k^2+x_k^4)\ge 3(1+x_k+x_k^2+x_k^3+x_k^4)$, which implies $$\frac{1+x_k^2+x_k^4}{1+x_k+x_k^2+x_k^3+x_k^4}\ge \frac{3}{5}.$$ By the rearrangement inequality (the sum of products in any order is greater than or equal to the sum in reverse order), $$\sum _{k=1}^n\frac{1+x_k^2+x_k^4}{1+x_{k+1}+x_{k+1}^2+x_{k+1}^3+x_{k+1}^4}\ge \sum _{k=1}^n\frac{1+x_k^2+x_k^4}{1+x_k+x_k^2+x_k^3+x_k^4}\ge \frac{3}{5}n.$$

(2) The maximum value is $n^4+n^2+n-1+\frac{n-1}{n^5-1}$. On one hand, taking $x_1=n$ and $x_2=x_3=\cdots =x_n=0$, we have $$\sum _{k=1}^n\frac{1+x_k^2+x_k^4}{1+x_{k+1}+x_{k+1}^2+x_{k+1}^3+x_{k+1}^4}=n^4+n^2+n-1+\frac{n-1}{n^5-1}.$$ On the other hand, for any $1\le k\le n$, let $a_k=1+x_k^2+x_k^4$ and $b_k=1+x_k+x_k^2+x_k^3+x_k^4$, with $b_{n+1}=b_1$. Since $a_k\ge 1$ and $\frac{1}{b_{k+1}}\le 1$, it follows that $(a_k-1)\left(\frac{1}{b_{k+1}}-1\right)\le 0$, and thus $$\frac{a_k}{b_{k+1}}\le a_k+\frac{1}{b_{k+1}}-1.$$ Therefore, $$\sum _{k=1}^n\frac{a_k}{b_{k+1}}\le \sum _{k=1}^n{a}_k+\sum _{k=1}^n\frac{1}{b_{k+1}}-n.$$ Let $f(x)=1+x^2+x^4$ for $x\ge 0$, and $g(x)=\frac{1}{1+x+x^2+x^3+x^4}$ for $x\ge 0$. It is clear that $f(x)$ is convex on $[0,+\infty )$.

We now prove that $g(x)$ is also convex on $[0,+\infty )$. Indeed, $$\begin{array}{rl}{g}^{\prime }(x)& =\frac{-4x^5+5x^4-1}{(x^5-1{)}^2},\\ g^{\prime \prime }(x)& =\frac{10(x^5-1)x^3(x-1{)}^2(2x^2+x+2)(x^2-1)}{(x^5-1{)}^4}\\ & =\frac{10x^3(x+1)(2x^2+x+2)}{(1+x+x^2+x^3+x^4{)}^2}\ge 0.\end{array}$$ It is well-known that if $\phi (x)$ is convex, then for any $a,b\ge 0$, $$\phi (a)+\phi (b)\le \phi (a+b)+\phi (0).$$ Applying this inequality repeatedly, we obtain $$\begin{array}{rl}& \sum _{k=1}^n{a}_k+\sum _{k=1}^n\frac{1}{b_{k+1}}-n\\ \le & \sum _{k=1}^{n}f(x_k)+\sum _{k=1}^{n}g(x_k)-n\\ \le & f\left(\sum _{k=1}^n{x}_k\right)+g\left(\sum _{k=1}^n{x}_k\right)+(n-1)(f(0)+g(0))-n\\ =& n^4+n^2+n-1+\frac{n-1}{n^5-1}.\end{array}$$