Example 2 For any real number sequence $\left\{x_{n}\right\}$, define the sequence $\left\{y_{n}\right\}:$
$$
y_{1}=x_{1}, y_{n+1}=x_{n+1}-\left(\sum_{i=1}^{n} x_{i}^{2}\right)^{\frac{1}{2}}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$

Find the smallest positive number $\lambda$, such that for any real number sequence $\left\{x_{n}\right\}$ and all positive integers $m$, we have
$$
\frac{1}{m} \sum_{i=1}^{m} x_{i}^{2} \leqslant \sum_{i=1}^{m} \lambda^{m-i} y_{i}^{2} .
$$
【Analysis】First estimate the upper bound of $\lambda$ from the limit perspective, then try to construct a recurrence relation to solve it.
First, prove that $\lambda \geqslant 2$.
In fact, start from simple and special cases.
Take $x_{1}=1, x_{n}=\sqrt{2^{n-2}}(n \geqslant 2)$. Then $y_{1}=1, y_{n}=0(n \geqslant 2)$.
Substitute into equation (1) to get
$$
\begin{array}{l}
\frac{1}{m} \cdot 2^{m-1} \leqslant \lambda^{m-1} \\
\Rightarrow \frac{2}{\lambda} \leqslant \sqrt[m-1]{m} .
\end{array}
$$

When $m \rightarrow \infty$, $\frac{2}{\lambda} \leqslant 1 \Rightarrow \lambda \geqslant 2$.
Next, use mathematical induction to prove: for sequences $\left\{x_{n}\right\}$ and $\left\{y_{n}\right\}$ that satisfy the conditions and for all positive integers $m$, equation (1) holds. When $m=1$, $x_{1}^{2}=y_{1}^{2} \leqslant y_{1}^{2}$, the conclusion holds.
Assume the conclusion holds for $m=k$.
Then when $m=k+1$, let
$$
\begin{array}{l}
f(k+1)=\sum_{i=1}^{k+1} 2^{k+1-i} y_{i}^{2}-\frac{1}{k+1} \sum_{i=1}^{k+1} x_{i}^{2}, \\
f(k)=\sum_{i=1}^{k} 2^{k-i} y_{i}^{2}-\frac{1}{k} \sum_{i=1}^{k} x_{i}^{2} .
\end{array}
$$

Thus, $f(k+1)-f(k)$
$$
\geqslant y_{k+1}^{2}+\frac{1}{k} \sum_{i=1}^{k} x_{i}^{2}-\frac{1}{k+1} x_{k+1}^{2}+f(k) \text {. }
$$

By the given conditions and the Cauchy-Schwarz inequality, we get
$$
\begin{array}{l}
x_{k+1}^{2}=\left(y_{k+1}+\sqrt{x_{1}^{2}+x_{2}^{2}+\cdots+x_{k}^{2}}\right)^{2} \\
\leqslant\left((k+1) y_{k+1}^{2}+\frac{k+1}{k} \sum_{i=1}^{k} x_{i}^{2}\right)^{2}\left(\frac{1}{k+1}+\frac{k}{k+1}\right) \\
\leqslant(k+1) y_{k+1}^{2}+\left(1+\frac{1}{k}\right) \sum_{i=1}^{k} x_{i}^{2} .
\end{array}
$$

By the induction hypothesis, we get
$$
f(k+1) \geqslant f(k) \geqslant 0 \text {. }
$$

Therefore, the conclusion holds when $m=k+1$.
In summary, the minimum value of the positive number $\lambda$ is 2.