Example 3.15 (2007 Girls' Mathematical Olympiad) Let $n>3$ be an integer, and let $a_{1}, a_{2}, \cdots, a_{n}$ be non-negative real numbers satisfying $a_{1}+a_{2}+\cdots+a_{n}=2$. Find the minimum value of
$$\frac{a_{1}}{a_{2}^{2}+1}+\frac{a_{2}}{a_{3}^{2}+1}+\cdots+\frac{a_{n}}{a_{1}^{2}+1}.$$
This problem first appeared in Mathematical Reflections, as a generalization of a 2003 Bulgarian competition problem (the original problem was for 3 variables), and was later selected as a 2007 Girls' Mathematical Olympiad problem. It is said that few people solved this problem that year, indicating its considerable difficulty. The challenge lies in the fact that using Cauchy, adjustments, or directly applying local inequalities for magnification are all difficult to achieve. First, we can easily guess the answer to be $\frac{3}{2}$, with equality holding if and only if $a_{1}=a_{2}=1, a_{3}=a_{4}=\cdots=a_{n}=0$ and its cyclic permutations. In this case, $\frac{a_{1}}{a_{2}^{2}+1}=\frac{1}{2}, \frac{a_{2}}{a_{3}^{2}+1}=1$, and their positions are unequal, so we consider splitting these two terms. For terms of the form $\frac{x}{y+1}$, a commonly used method is to write them as $x-\frac{x y}{y+1}$. For this problem:
$$\begin{array}{l}
\frac{a_{1}}{a_{2}^{2}+1}+\frac{a_{2}}{a_{3}^{2}+1}+\cdots+\frac{a_{n}}{a_{1}^{2}+1}= \\
a_{1}+a_{2}+\cdots+a_{n}-\left(\frac{a_{1} a_{2}^{2}}{a_{2}^{2}+1}+\frac{a_{2} a_{3}^{2}}{a_{3}^{2}+1}+\cdots+\frac{a_{n} a_{1}^{2}}{a_{1}^{2}+1}\right) \leqslant \\
2-\frac{1}{2}\left(a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n} a_{1}\right)
\end{array}$$

Thus, we need to prove
$$4\left(a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n} a_{1}\right) \leqslant\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}$$

This is a common inequality that has appeared in many competition books, and the methods in these books generally use mathematical induction. Here, we provide a clever proof. Without loss of generality, assume $a_{1}=\max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$, then
$$\begin{aligned}
4\left(a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n} a_{1}\right) \leqslant & 4 a_{1}\left(a_{2}+a_{4}+\cdots+a_{n}\right)+4 a_{2} a_{3}+4 a_{3} a_{4} \leqslant \\
& 4\left(a_{1}+a_{3}\right)\left(a_{2}+a_{4}+\cdots+a_{4}\right) \leqslant \\
& \left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}
\end{aligned}$$

Therefore, $\square$
$$\frac{a_{1}}{a_{2}^{2}+1}+\frac{a_{2}}{a_{3}^{2}+1}+\cdots+\frac{a_{n}}{a_{1}^{2}+1} \geqslant \frac{3}{2}$$

Equality holds if and only if $a_{1}=a_{2}=1, a_{3}=a_{4}=\cdots=a_{n}=0$ and its cyclic permutations.