Problem 1334

The answer is $\frac{3}{2}$.

The given problem is equivalent to finding the minimum value of
$$\begin{aligned} m &= 2 - \left( \frac{a_1}{a_2^2+1} + \frac{a_2}{a_3^2+1} + \cdots + \frac{a_n}{a_1^2+1} \right) \\ &= \left( a_1 - \frac{a_1}{a_2^2+1} \right) + \left( a_2 - \frac{a_2}{a_3^2+1} \right) + \cdots + \left( a_n - \frac{a_n}{a_1^2+1} \right) \\ &= \frac{a_1 a_2^2}{a_2^2+1} + \frac{a_2 a_3^2}{a_3^2+1} + \cdots + \frac{a_n a_1^2}{a_1^2+1}. \end{aligned}$$
Since $a_i^2 + 1 \ge 2a_i$, we have
$$m \le \frac{a_1 a_2 + a_2 a_3 + \cdots + a_n a_1}{2}.$$
Our result follows from the following well-known fact:
$$\begin{aligned} & f(a_1, \cdots, a_n) \\ &= (a_1 + \cdots + a_n)^2 - 4(a_1 a_2 + a_2 a_3 + \cdots + a_n a_1) \\ &\ge 0 \end{aligned} \qquad \textcircled{1}$$
for integers $n \ge 4$ and nonnegative real numbers $a_1, a_2, \cdots, a_n$.

To prove this fact, we use induction on $n$. For $n=4$, $\textcircled{1}$ becomes
$$\begin{aligned} & f(a_1, a_2, a_3, a_4) \\ &= (a_1 + a_2 + a_3 + a_4)^2 - 4(a_1 a_2 + a_2 a_3 + a_3 a_4 + a_4 a_1) \\ &= (a_1 + a_2 + a_3 + a_4)^2 - 4(a_1 + a_3)(a_2 + a_4), \end{aligned}$$
which is nonnegative by the AM-GM inequality.

Assume that $\textcircled{1}$ is true for $n=k$ for some integer $k \ge 4$.

Consider the case when $n = k + 1$. By (cyclic) symmetry in $\textcircled{1}$, we may assume that $a_{k+1} = \min\{a_1, a_2, \dots, a_{k+1}\}$. By the induction hypothesis, it suffices to show that
$$D = f(a_1, \dots, a_k, a_{k+1}) - f(a_1, \dots, a_{k-1}, a_k + a_{k+1}) \ge 0.$$
Note that
$$\begin{aligned} \frac{D}{4} &= -(a_1a_2 + \cdots + a_k a_{k+1} + a_{k+1}a_1) + [a_1a_2 + \cdots \\ & \qquad \qquad + a_{k-1}(a_k + a_{k+1}) + (a_k + a_{k+1})a_1] \\ &= a_{k-1}a_{k+1} + a_k a_1 - a_k a_{k+1} \\ &= a_{k-1}a_{k+1} + (a_1 - a_{k+1})a_k \\ &\ge 0, \end{aligned}$$
completing our proof.

From the above result, we can get
$$\begin{aligned} & \frac{1}{2}(a_1a_2 + a_2a_3 + \cdots + a_n a_1) \\ \le & \frac{1}{8}(a_1 + a_2 + \cdots + a_n)^2 \\ = & \frac{1}{8} \times 2^2 \\ = & \frac{1}{2}, \end{aligned}$$
therefore
$$\frac{a_1 a_2^2}{a_2^2 + 1} + \frac{a_2 a_3^2}{a_3^2 + 1} + \cdots + \frac{a_n a_1^2}{a_1^2 + 1} \le \frac{1}{2},$$
that is $m \ge \frac{3}{2}$. When $a_1 = a_2 = 1$ and $a_3 = \cdots = a_n = 0$, $m = \frac{3}{2}$ holds.