China Girls' Mathematical Olympiad

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

The given problem is equivalent to finding the minimum value of $$\begin{array}{rl}m& =2-\left(\frac{a_1}{a_2^2+1}+\frac{a_2}{a_3^2+1}+\cdots +\frac{a_n}{a_1^2+1}\right)\\ & =\left(a_1-\frac{a_1}{a_2^2+1}\right)+\left(a_2-\frac{a_2}{a_3^2+1}\right)+\cdots +\left(a_n-\frac{a_n}{a_1^2+1}\right)\\ & =\frac{a_1{a}_2^2}{a_2^2+1}+\frac{a_2{a}_3^2}{a_3^2+1}+\cdots +\frac{a_n{a}_1^2}{a_1^2+1}.\end{array}$$ 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: $$\begin{array}{rl}& f(a_1,\cdots ,a_n)\\ & =(a_1+\cdots +a_n{)}^2-4(a_1{a}_2+a_2{a}_3+\cdots +a_n{a}_1)\\ & \ge 0\end{array}\stackrel{◯}{1}$$ 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 $n=4$, $\stackrel{◯}{1}$ becomes $$\begin{array}{rl}& f(a_1,a_2,a_3,a_4)\\ & =(a_1+a_2+a_3+a_4{)}^2-4(a_1{a}_2+a_2{a}_3+a_3{a}_4+a_4{a}_1)\\ & =(a_1+a_2+a_3+a_4{)}^2-4(a_1+a_3)(a_2+a_4),\end{array}$$ which is nonnegative by the AM-GM inequality.

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

Consider the case when $n=k+1$. By (cyclic) symmetry in $\stackrel{◯}{1}$, 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 $$\begin{array}{rl}\frac{D}{4}& =-(a_1{a}_2+\cdots +a_k{a}_{k+1}+a_{k+1}{a}_1)+[a_1{a}_2+\cdots \\ & +a_{k-1}(a_k+a_{k+1})+(a_k+a_{k+1})a_1]\\ & =a_{k-1}{a}_{k+1}+a_k{a}_1-a_k{a}_{k+1}\\ & =a_{k-1}{a}_{k+1}+(a_1-a_{k+1})a_k\\ & \ge 0,\end{array}$$ completing our proof.

From the above result, we can get $$\begin{array}{rl}& \frac{1}{2}(a_1{a}_2+a_2{a}_3+\cdots +a_n{a}_1)\\ \le & \frac{1}{8}(a_1+a_2+\cdots +a_n{)}^2\\ =& \frac{1}{8}\times 2^2\\ =& \frac{1}{2},\end{array}$$ 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.