Example 5 (2002 China National Team Training Selection Test for IMO) Let $a_{1}=\frac{1}{4}, a_{n}=\frac{1}{4}\left(1+a_{n-1}\right)^{2}, n \geqslant 2$. Find the smallest real number $\lambda$, such that for any non-negative real numbers $x_{1}, x_{2}, \cdots, x_{2002}$, we have
$$
\sum_{k=1}^{2002} A_{k} \leqslant \lambda a_{2002} .
$$

where $A_{k}=\frac{x_{k}-k}{\left(x_{k}+\cdots+x_{2002}+\frac{k(k-1)}{2}+1\right)^{2}}, k \geqslant 1$.
Proof According to the problem, the original problem is transformed into a chain of problems.
At this point, let $\delta(k)=\frac{1}{2} k(k-1)$, and consider a series of lemmas:
Lemma 1 For any real numbers $a \geqslant 0, c>0, b>0$, the function $f(x)=\frac{a}{x+b}+\frac{x-c}{(x+b)^{2}}$. When $x=\frac{(1-a) b+2 c}{1+a}$, it reaches the maximum value $\frac{1}{4} \cdot \frac{(1+a)^{2}}{b+c}$.
In fact, let $y=\frac{1}{x+b}$, then
$$
f(x)=-(b+c) y^{2}+(1+a) y=-(b+c)\left(y-\frac{1}{2} \cdot \frac{1+a}{b+c}\right)^{2}+\frac{1}{4} \cdot \frac{(1+a)^{2}}{b+c} \text {. }
$$

Thus, when $y=\frac{1}{2} \cdot \frac{1+a}{b+c}$, then $x=\frac{(1-a) b+2 c}{1+a}$,
$$
f(x)_{\min }=\frac{1}{4} \cdot \frac{(1+a)^{2}}{b+c} \text {. }
$$

Lemma 2 Let $a_{1}=\frac{1}{4}, a_{n}=\frac{1}{4}\left(1+a_{n-1}\right)^{2}, n \geqslant 2$. Then $a_{n}$ satisfies $0<a_{n}<1$.
Lemma 3 For any $n \geqslant 1, \sum_{k=1}^{n} A_{k} \leqslant \frac{1}{\delta(n+1)+1} a_{n}$, and equality can be achieved.
In fact, by Lemma 1, we have
$$
\frac{x_{1}-1}{\left(x_{1}+\cdots+x_{n}+1\right)^{2}} \leqslant \frac{1}{4} \cdot \frac{1}{x_{2}+\cdots+x_{n}+2}=\frac{a_{1}}{x_{2}+\cdots+x_{n}+2},
$$

and when $x_{1}=x_{2}+\cdots+x_{n}+3$, it reaches the maximum value
$$
\begin{aligned}
& \frac{a_{1}}{x_{2}+\cdots+x_{n}+2} \cdot \frac{a_{1}}{x_{2}+\cdots+x_{n}+2}+\frac{x_{2}-2}{\left(x_{2}+\cdots+x_{n}+2\right)^{2}} \\
\leqslant & \frac{1}{4} \cdot \frac{\left(1+a_{1}\right)^{2}}{x_{3}+\cdots+x_{n}+4}=\frac{a_{2}}{x_{3}+\cdots+x_{n}+4},
\end{aligned}
$$

and when $x_{2}=\frac{\left(1-a_{1}\right)\left(x_{3}+\cdots+x_{n}+4\right)+4}{1+a_{1}}$, it reaches the maximum value $\frac{a_{2}}{x_{3}+\cdots+x_{n}+4}$.
...
$$
\begin{aligned}
& \frac{a_{n-2}}{x_{n-1}+x_{n}+\delta(n-1)+1}+\frac{x_{n-1}-(n-1)}{\left(x_{n-1}+x_{n}+\delta(n-1)+1\right)^{2}} \\
\leqslant & \frac{1}{4} \cdot \frac{\left(1+a_{n-2}\right)^{2}}{x_{n}+\delta(n)+1}=\frac{a_{n-1}}{x_{n}+\delta(n)+1},
\end{aligned}
$$

and when $x_{n-1}=\frac{\left[\left(1-a_{n-2}\right)\left(x_{n}+\delta(n-1)+1\right)+2(n-1)\right]}{1+a_{n-2}}$
it reaches the maximum value $\frac{a_{n-1}}{x_{n}+\delta(n)+1}$.
$$
\begin{aligned}
& \frac{a_{n-1}}{x_{n}+\delta(n)+1}+\frac{x_{n}-n}{\left(x_{n}+\delta(n)+1\right)^{2}} \\
\leqslant & \frac{1}{4} \cdot \frac{\left(1+a_{n-1}\right)^{2}}{\delta(n+1)+1}=\frac{a_{n}}{\delta(n+1)+1},
\end{aligned}
$$

and when $x_{n}=\frac{\left(1-a_{n-1}\right)(\delta(n)+1)+2 n}{1+a_{n-1}}$, it reaches the maximum value $\frac{a_{n}}{\delta(n+1)+1}$.
By (1), (2), ..., adding them up, we get
$$
\sum_{k=1}^{n} A_{k} \leqslant \frac{1}{\delta(n+1)+1} a_{n},
$$

and when $x_{n}=\frac{\left(1-a_{n-1}\right)(\delta(n)+1)+2 n}{1+a_{n-1}}$,
$$
\begin{array}{l}
x_{n-1}=\frac{\left(1-a_{n-2}\right)\left(x_{n}+\delta(n-1)+1\right)+2(n-1)}{1+a_{n-2}}, \\
\cdots \\
x_{2}=\frac{\left(1-a_{1}\right)\left(x_{3}+\cdots+x_{n}+4\right)+4}{1+a_{1}}, \\
x_{1}=x_{2}+\cdots+x_{n}+3
\end{array}
$$

equality holds.
By Lemma 3, we get
$$
\lambda=\frac{1}{\delta(2003)+1}=\frac{1}{2003 \times 1001+1} .
$$