Example 2 For any real sequence $\left\{x_{n}\right\}$, define the sequence $\left\{y_{n}\right\}$ as follows:
$$
y_{1}=x_{1}, y_{n+1}=x_{n+1}-\left(\sum_{i=1}^{n} x_{i}^{2}\right)^{\frac{1}{2}}(n \geqslant 1) \text {. }
$$

Find the smallest positive number $\lambda$, such that for any real sequence $\left\{x_{n}\right\}$ and all positive integers $m$, we have
$$
\frac{1}{m} \sum_{i=1}^{m} x_{i}^{2} \leqslant \sum_{i=1}^{m} \lambda^{m-i} y_{i}^{2} \text {. }
$$
(2014, China National Training Team Test)
【Analysis】First, conduct a mathematical experiment.
When $m=1$, we have $x_{1}^{2} \leqslant y_{1}^{2}=x_{1}^{2}$, the inequality is obviously
true.
When $m=2$, the original inequality is equivalent to
$\frac{x_{1}^{2}+x_{2}^{2}}{2} \leqslant \lambda y_{1}^{2}+y_{2}^{2}=\lambda x_{1}^{2}+\left(x_{1}^{2}+x_{2}^{2}-2 x_{2}\left|x_{1}\right|\right)$
$\Leftrightarrow(2 \lambda+1) x_{1}^{2}+x_{2}^{2}-4 x_{2}\left|x_{1}\right| \geqslant 0$
$\Leftrightarrow(2 \lambda-3) x_{1}^{2}+\left(2\left|x_{1}\right|-x_{2}\right)^{2} \geqslant 0$.
The above inequality holds for any real numbers $x_{1}, x_{2}$, and

the necessary and sufficient condition is $\lambda \geqslant \frac{3}{2}$.
When $m=3$, the original inequality is equivalent to
$\frac{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}{3} \leqslant \lambda^{2} y_{1}^{2}+\lambda y_{2}^{2}+y_{3}^{2}$
$=\lambda^{2} x_{1}^{2}+\lambda\left(x_{2}-\left|x_{1}\right|\right)^{2}+\left(x_{3}-\sqrt{x_{1}^{2}+x_{2}^{2}}\right)^{2}$
$\Leftrightarrow\left(3 \lambda^{2}+3 \lambda+2\right) x_{1}^{2}+(3 \lambda+2) x_{2}^{2}+2 x_{3}^{2}-$
$6 \lambda x_{2}\left|x_{1}\right|-6 x_{3} \sqrt{x_{1}^{2}+x_{2}^{2}} \geqslant 0$
$\Leftrightarrow 2\left(x_{3}-\frac{3}{2} \sqrt{x_{1}^{2}+x_{2}^{2}}\right)^{2}+$
$\left(3 \lambda^{2}+3 \lambda-\frac{5}{2}\right) x_{1}^{2}+\left(3 \lambda-\frac{5}{2}\right) x_{2}^{2}-$
$6 \lambda x_{2}\left|x_{1}\right| \geqslant 0$
$\Leftrightarrow 2\left(x_{3}-\frac{3}{2} \sqrt{x_{1}^{2}+x_{2}^{2}}\right)^{2}+$
$\left(3 \lambda-\frac{5}{2}\right)\left(x_{2}-\frac{6 \lambda}{6 \lambda-5}\left|x_{1}\right|\right)^{2}+$
$\frac{36 \lambda^{3}-30 \lambda^{2}-60 \lambda+25}{12 \lambda-10} \geqslant 0$.

The above inequality holds for any real numbers $x_{1}, x_{2}, x_{3}$,
and the necessary and sufficient condition is
$\left\{\begin{array}{l}6 \lambda-5>0, \\ 36 \lambda^{3}-30 \lambda 20\end{array}\right.$
$\left\{36 \lambda^{3}-30 \lambda^{2}-60 \lambda+25 \geqslant 0\right.$.
This system of inequalities is difficult to solve.
When $m \geqslant 3$, the discussion using the above method will be more complex.

Therefore, consider the extreme case $y_{2}=y_{3}=\cdots=0$.
In fact, when $x_{1}=1$, then $y_{1}=1$;
when $y_{2}=0$, $x_{2}=1$;
when $y_{3}=0$, $x_{3}=\sqrt{2}$;
when $y_{4}=0$, $x_{4}=2$;
when $y_{5}=0$, $x_{5}=2 \sqrt{2}$;
when $y_{6}=0$, $x_{6}=4$;
Summarizing the pattern, when $n \geqslant 2$ and $y_{n}=0$,
$x_{n}=2^{\frac{n}{2}-1}$.
Thus, $\frac{1}{m} \sum_{i=1}^{m} x_{i}^{2}=\frac{1}{m}\left(1+\sum_{i=2}^{m} 2^{i-2}\right)=\frac{2^{m-1}}{m}$
$\leqslant \sum_{i=1}^{m} \lambda^{m-i} y_{i}^{2}=\lambda^{m-1}$,

i.e., $\lambda \geqslant \frac{2}{m^{\frac{1}{1-1}}}$,
for any positive integer $m \geqslant 2$.
Lemma 1 If the positive integer $m \geqslant 2$, then
$m^{\frac{1}{n-1}}>1$, and $\lim _{m \rightarrow \infty} m^{\frac{1}{m-1}}=1$.
Proof of Lemma 1 Let $\alpha_{m}=\sqrt[m-1]{m}-1$.
Clearly, when $m \geqslant 3$, $\alpha_{m}>0$.
Notice that,
$m=\left(1+\alpha_{m}\right)^{m-1}$
$=1+(m-1) \alpha_{m}+\frac{(m-1)(m-2)}{2} \alpha_{m}^{2}+\cdots$
$>1+\frac{(m-1)(m-2)}{2} \alpha_{m}^{2}$,

i.e., $0<\alpha_{m}<\sqrt{\frac{2}{m-2}}(m \geqslant 3)$.
Since $\lim _{m \rightarrow \infty} \sqrt{\frac{2}{m-2}}=0$, combining the squeeze theorem, we get
$\lim _{m \rightarrow \infty} \alpha_{m}=0$.
Lemma 1 is proved.
Using Lemma 1 and equation (1), we get $\lambda \geqslant 2$.
Next, we use mathematical induction to prove: when $\lambda=2$, the conclusion of the problem
holds, i.e., for sequences $\left\{x_{n}\right\}$ and $\left\{y_{n}\right\}$ satisfying the conditions
and all positive integers $m$, we have
$\frac{1}{m} \sum_{i=1}^{m} x_{i}^{2} \leqslant \sum_{i=1}^{m} 2^{m-i} y_{i}^{2}$.
From the previous mathematical experiment, when (2)

is obviously true.
Assume that when $m=k$, equation (2) holds,
then when $m=k+1$, let
$f(k+1)=\sum_{i=1}^{k+1} 2^{k+1-i} y_{i}^{2}-\frac{1}{k+1} \sum_{i=1}^{k+1} x_{i}^{2}$,
$f(k)=\sum_{i=1}^{k} 2^{k-i} y_{i}^{2}-\frac{1}{k} \sum_{i=1}^{k} x_{i}^{2}$.
Then $f(k+1)-2 f(k)$
$=y_{k+1}^{2}+\frac{2}{k} \sum_{i=1}^{k} x_{i}^{2}-\frac{1}{k+1} \sum_{i=1}^{k+1} x_{i}^{2}$
$=y_{k+1}^{2}+\frac{1}{k} \sum_{i=1}^{k} x_{i}^{2}-\frac{1}{k+1} x_{k+1}^{2}+$
$\left(\frac{1}{k}-\frac{1}{k+1}\right) \sum_{i=1}^{k} x_{i}^{2}$
$\geqslant y_{k+1}^{2}+\frac{1}{k} \sum_{i=1}^{k} x_{i}^{2}-\frac{1}{k+1} x_{k+1}^{2}$.
Thus, it suffices to prove
$y_{k+1}^{2}+\frac{1}{k} \sum_{i=1}^{k} x_{i}^{2}-\frac{1}{k+1} x_{k+1}^{2} \geqslant 0$
(3)
By the conditions and the Cauchy-Schwarz inequality, we get
$(k+1)\left(y_{k+1}^{2}+\frac{1}{k} \sum_{i=1}^{k} x_{i}^{2}\right)$
$=\left((k+1) y_{k+1}^{2}+\frac{k+1}{k} \sum_{i=1}^{k} x_{i}^{2}\right)\left(\frac{1}{k+1}+\frac{k}{k+1}\right)$
$\geqslant\left(y_{k+1}+\sqrt{x_{1}^{2}+x_{2}^{2}+\cdots+x_{k}^{2}}\right)^{2}=x_{k+1}^{2}$,

i.e., equation (3) holds.
Thus, according to the induction hypothesis,
$f(k+1) \geqslant 2 f(k) \geqslant 0$.
By mathematical induction, equation (2) holds.
In summary, the minimum value of the positive number $\lambda$ is 2.
【Note】By the principle of extremality, to facilitate the calculation of $\sum_{i=1}^{m} \lambda^{m-i} y_{i}^{2}$,
consider the special case $y_{2}=y_{3}=\cdots=0$, and construct an example to show $\lambda \geqslant 2$; then use mathematical induction to prove that when $\lambda=2$, the original inequality holds. The inequality
$y_{k+1}^{2}+\frac{1}{k} \sum_{i=1}^{k} x_{i}^{2}-\frac{1}{k+1} x_{k+1}^{2}