33. 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 $\frac{a_{1}}{a_{2}^{2}+1}+\frac{a_{2}}{a_{3}^{2}+1}+\cdots+\frac{a_{n}}{a_{1}^{2}+1}$. 2007 China Girls' Mathematical Olympiad
33. From $a_{1}+a_{2}+\cdots+a_{n}=2$, the problem is equivalent to finding the maximum value of:
$$\begin{array}{l}
a_{1}-\frac{a_{1}}{a_{2}^{2}+1}+a_{2}-\frac{a_{2}}{a_{3}^{2}+1}+\cdots+a_{n}-\frac{a_{n}}{a_{1}^{2}+1}= \\
\frac{a_{1} a_{2}^{2}}{a_{2}^{2}+1}+\frac{a_{2} a_{3}^{2}}{a_{3}^{2}+1}+\cdots+\frac{a_{n} a_{1}^{2}}{a_{1}^{2}+1} \leqslant \\
\frac{1}{2}\left(a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n} a_{1}\right)
\end{array}$$

The last inequality in the above expression holds because $x^{2}+1 \geqslant 2 x$. When $x>0, y \geqslant 0$, $\frac{1}{2 x} \geqslant \frac{1}{x^{2}+1}$, $\frac{y x^{2}}{2 x} \geqslant \frac{y x^{2}}{x^{2}+1}$, i.e., $\frac{y x^{2}}{x^{2}+1} \leqslant \frac{x y}{2}$; when $x=0$, the above inequality also holds.

Lemma: If $a_{1}, a_{2}, \cdots, a_{n} \geqslant 0, n \geqslant 4$, then $4\left(a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n} a_{1}\right) \leqslant\left(a_{1}+\right.$ $\left.a_{2}+\cdots+a_{n}\right)^{2}$

Let $f\left(a_{1}, a_{2}, \cdots, a_{n}\right)=4\left(a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n} a_{1}\right)-\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}$.
We will prove by mathematical induction that
$$f\left(a_{1}, a_{2}, \cdots, a_{n}\right) \leqslant 0$$

When $n=4$, the inequality (1) is equivalent to $4\left(a_{1}+a_{3}\right)\left(a_{2}+a_{4}\right) \leqslant\left(a_{1}+a_{2}+a_{3}+a_{4}\right)^{2}$. By the AM-GM inequality, the inequality holds.

Assume the inequality (1) holds for $n=k(k \geqslant 4)$. For $n=k+1$, without loss of generality, assume $a_{k}=\min \left\{a_{1}\right.$, $\left.a_{2}, \cdots, a_{k}, a_{k+1}\right\}$, then
$$\begin{array}{l}
f\left(a_{1}, a_{2}, \cdots, a_{k}, a_{k+1}\right)-f\left(a_{1}, a_{2}, \cdots, a_{k-1}, a_{k}+a_{k+1}\right)= \\
4\left[a_{k-1} a_{k}+a_{k} a_{k+1}+a_{1} a_{k+1}-a_{k-1}\left(a_{k}+a_{k+1}\right)-\left(a_{k}+a_{k+1}\right) a_{1}\right]= \\
-4\left[\left(a_{k-1}-a_{k}\right) a_{k+1}+a_{1} a_{k}\right] \leqslant 0
\end{array}$$

i.e.,
$$f\left(a_{1}, a_{2}, \cdots, a_{k}, a_{k+1}\right) \leqslant f\left(a_{1}, a_{2}, \cdots, a_{k-1}, a_{k}+a_{k+1}\right)$$

By the induction hypothesis, the right-hand side of the above inequality is $\leqslant 0$. Thus, the inequality (1) holds for $n=k+1$. The lemma is proved.
By the lemma, we have:
$$\begin{array}{l}
\frac{1}{2}\left(a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n} a_{1}\right) \leqslant \\
\frac{1}{8}\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}= \\
\frac{1}{8} \times 2^{2}=\frac{1}{2}
\end{array}$$

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

i.e.,
$$\frac{a_{1}}{a_{2}^{2}+1}+\frac{a_{2}}{a_{3}^{2}+1}+\cdots+\frac{a_{n}}{a_{1}^{2}+1} \geqslant \frac{3}{2}$$

When $a_{1}=a_{2}=1, a_{3}=\cdots=a_{n}=0$, the above inequality can be an equality. Hence, the minimum value is $\frac{3}{2}$.