3. Let integer $n(n>3)$, non-negative real numbers $a_{1}, a_{2}$, $\cdots, a_{n}$ satisfy $a_{1}+a_{2}+\cdots+a_{n}=2$.
Find the minimum value of $\frac{a_{1}}{a_{2}^{2}+1}+\frac{a_{2}}{a_{3}^{2}+1}+\cdots+\frac{a_{n}}{a_{1}^{2}+1}$.
(Provided by Zhu Huawei)
3. From $a_{1}+a_{2}+\cdots+a_{n}=2$, we know the problem is equivalent to finding the maximum value of
$$
\begin{array}{l}
a_{1}-\frac{a_{1}}{a_{2}^{2}+1}+a_{2}-\frac{a_{2}}{a_{3}^{2}+1}+\cdots+a_{n}-\frac{a_{n}}{a_{1}^{2}+1} \\
=\frac{a_{1} a_{2}^{2}}{a_{2}^{2}+1}+\frac{a_{2} a_{3}^{2}}{a_{3}^{2}+1}+\cdots+\frac{a_{n} a_{1}^{2}}{a_{1}^{2}+1}
\end{array}
$$

Since $x^{2}+1 \geqslant 2 x$, for $x>0, y \geqslant 0$,
$$
\frac{1}{2 x} \geqslant \frac{1}{x^{2}+1}, \frac{y x^{2}}{2 x} \geqslant \frac{y x^{2}}{x^{2}+1} \text {, }
$$

i.e., $\frac{y x^{2}}{x^{2}+1} \leqslant \frac{1}{2} x y$;
When $x=0$, the above inequality also holds.
Thus, $\frac{a_{1} a_{2}^{2}}{a_{2}^{2}+1}+\frac{a_{2} a_{3}^{2}}{a_{3}^{2}+1}+\cdots+\frac{a_{n} a_{1}^{2}}{a_{1}^{2}+1}$
$$
\leqslant \frac{1}{2}\left(a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n} a_{1}\right) .
$$

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

Proof of the lemma: Let
$$
\begin{array}{l} 
f\left(a_{1}, a_{2}, \cdots, a_{n}\right) \\
= 4\left(a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n-1} a_{n}+a_{n} a_{1}\right)- \\
\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2} .
\end{array}
$$

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
$$
4\left(a_{1}+a_{3}\right)\left(a_{2}+a_{4}\right) \leqslant\left(a_{1}+a_{2}+a_{3}+a_{4}\right)^{2} \text {. }
$$

By the AM-GM inequality, the proposition holds.
Assume the inequality (1) holds for $n=k(k \geqslant 4)$.
For $n=k+1$, without loss of generality, assume
$$
a_{k}=\min \left\{a_{1}, a_{2}, \cdots, a_{k+1}\right\} \text {, }
$$

then $f\left(a_{1}, a_{2}, \cdots, a_{k+1}\right)-$
$$
\begin{array}{l} 
f\left(a_{1}, a_{2}, \cdots, a_{k-1}, a_{k}+a_{k+1}\right) \\
= 4\left[a_{k-1} a_{k}+a_{k} a_{k+1}+a_{1} a_{k+1}-\right. \\
\left.a_{k-1}\left(a_{k}+a_{k+1}\right)-\left(a_{k}+a_{k+1}\right) a_{1}\right] \\
=-4\left[\left(a_{k-1}-a_{k}\right) a_{k+1}+a_{1} a_{k}\right] \leqslant 0,
\end{array}
$$

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

By the induction hypothesis, the right-hand side is less than or equal to 0, i.e., when $n=k+1$, the inequality (1) holds.
Returning to the original problem.
By the lemma,
$$
\begin{array}{l}
\frac{1}{2}\left(a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n} a_{1}\right) \\
\leqslant \frac{1}{8}\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}=\frac{1}{8} \times 2^{2}=\frac{1}{2} .
\end{array}
$$

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

When $a_{1}=a_{2}=1, a_{3}=\cdots=a_{n}=0$, the above inequality holds with equality.
Therefore, the minimum value sought is $\frac{3}{2}$.