CGMO, 2024

Proof. The minimum value is $\frac{3}{5}n$.
On one hand, taking $x_1 = x_2 = \dots = x_n = 1$, 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 $x_2 = x_3 = \dots = x_n = 0$, 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,
$$\begin{align*} g'(x) &= \frac{-4x^5 + 5x^4 - 1}{(x^5 - 1)^2}, \\ g''(x) &= \frac{10(x^5 - 1)x^3(x - 1)^2(2x^2 + x + 2)(x^2 - 1)}{(x^5 - 1)^4} \\ &= \frac{10x^3(x + 1)(2x^2 + x + 2)}{(1 + x + x^2 + x^3 + x^4)^2} \ge 0. \end{align*}$$
It is well-known that if $\varphi(x)$ is convex, then for any $a, b \ge 0$,
$$\varphi(a) + \varphi(b) \le \varphi(a + b) + \varphi(0).$$
Applying this inequality repeatedly, we obtain
$$\begin{align*} & \sum_{k=1}^{n} a_k + \sum_{k=1}^{n} \frac{1}{b_{k+1}} - n \\ \le & \sum_{k=1}^{n} f(x_k) + \sum_{k=1}^{n} g(x_k) - n \\ \le & f\left(\sum_{k=1}^{n} x_k\right) + g\left(\sum_{k=1}^{n} x_k\right) + (n-1)(f(0) + g(0)) - n \\ = & n^4 + n^2 + n - 1 + \frac{n-1}{n^5-1}. \end{align*}$$