Question 1 Let two positive sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy:
(1) $a_{0}=1 \geqslant a_{1}$,
$a_{n}\left(b_{n-1}+b_{n+1}\right)=a_{n-1} b_{n-1}+a_{n+1} b_{n+1}(n \geqslant 1)$;
(2) $\sum_{i=0}^{n} b_{i} \leqslant n^{\frac{3}{2}}(n \geqslant 1)$.

Find the general term of $\left\{a_{n}\right\}$.
(Supplied by Li Weiguo)

Comment: This problem can be reduced to a recursive sequence problem, but the recursive method is quite novel, being a mixture of two sequences. If we regard $\left\{a_{n}\right\}$ as the main sequence (main variable), then $\left\{b_{n}\right\}$ can be seen as a given known sequence. In this way, if we separate the parameters, we can iterate in the usual way.

The key to solving the problem is to prove that $a_{1}=a_{0}$. The technique used is "counting twice": under the assumption that $a_{1}<a_{0}$, estimate the partial sum of a positive series from two perspectives to derive a contradiction.
Solution: From condition (1), we have
$$
a_{n}-a_{n+1}=\frac{b_{n-1}}{b_{n+1}}\left(a_{n-1}-a_{n}\right) \text {. }
$$

Thus, $a_{n}-a_{n+1}=\frac{b_{0} b_{1}}{b_{n} b_{n+1}}\left(a_{0}-a_{1}\right)$.
If $a_{1}=a_{0}=1$, then $a_{n}=1$.
Next, assume $a_{1}a_{0}-a_{n}=b_{0} b_{1}\left(a_{0}-a_{1}\right) \sum_{k=0}^{n-1} \frac{1}{b_{k} b_{k+1}} \text {. }
$$

Therefore, $\sum_{k=0}^{n} \frac{1}{b_{k} b_{k+1}}<\frac{a_{0}}{b_{0} b_{1}\left(a_{0}-a_{1}\right)}$.
On the other hand, let $x_{n}=\sqrt{b_{n} b_{n+1}}$, then
$x_{n} \leqslant \frac{b_{n}+b_{n+1}}{2}$.
Hence $\sum_{i=1}^{n} x_{i} \leqslant \sum_{i=1}^{n} \frac{b_{i}+b_{i+1}}{2}<\sum_{i=1}^{n+1} b_{i}$
$\leqslant(n+1)^{\frac{3}{2}} \leqslant 4 n^{\frac{3}{2}}$.
For any positive integer $k$, we have
$$
\begin{array}{l}
\frac{k}{\frac{1}{x_{k+1}^{2}}+\frac{1}{x_{k+2}^{2}}+\cdots+\frac{1}{x_{2 k}^{2}}} \leqslant \sqrt[k]{x_{k+1}^{2} \cdots x_{2 k}^{2}} \\
\leqslant\left(\frac{x_{k+1}+\cdots+x_{2 k}}{k}\right)^{2}<\frac{\left[4(2 k)^{\frac{3}{2}}\right]^{2}}{k^{2}} .
\end{array}
$$

Thus, $\frac{1}{x_{k+1}^{2}}+\frac{1}{x_{k+2}^{2}}+\cdots+\frac{1}{x_{2 k}^{2}} \geqslant \frac{1}{2^{7}}$.
Hence $\sum_{k=0}^{2^{n}} \frac{1}{b_{k} b_{k+1}}=\sum_{k=0}^{2^{n}} \frac{1}{x_{k}^{2}}=\sum_{k=0}^{n-1} \sum_{i=1}^{2^{k}} \frac{1}{x_{2^{2}+i}^{2^{2}}}$
$$
\geqslant \sum_{k=0}^{n-1} \frac{1}{2^{7}}=\frac{n}{2^{7}},
$$

which contradicts equation (2).
In conclusion, $a_{n}=1(n \geqslant 0)$.