Example 5 Let two sequences of positive numbers $\left\{a_{n}\right\},\left\{b_{n}\right\}$ 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 $\left\{a_{n}\right\}$.
From condition (1), we have $a_{n}-a_{n+1}=\frac{b_{n-1}}{b_{n+1}}\left(a_{n-1}-a_{n}\right)$, hence
$$a_{n}-a_{n+1}=\frac{b_{0} b_{1}}{b_{n} b_{n+1}}\left(a_{0}-a_{1}\right)$$

If $a_{1}=a_{0}=1$, then $a_{n}=1$. Below, we discuss $a_{1}a_{0}-a_{n}=b_{0} b_{1}\left(a_{0}-a_{1}\right) \sum_{k=0}^{n-1} \frac{1}{b_{k} b_{k+1}}$

This leads to $\quad \sum_{k=0}^{n} \frac{1}{b_{k} b_{k+1}}<\frac{a_{0}}{b_{0} b_{1}\left(a_{0}-a_{1}\right)}$.
Let $x_{n}=\sqrt{b_{n} b_{n+1}}$, then
$$\begin{aligned}
x_{1}+x_{2}+\cdots+x_{n} & \leqslant \frac{b_{1}+b_{2}}{2}+\frac{b_{2}+b_{3}}{2}+\cdots+\frac{b_{n}+b_{n+1}}{2} \\
& <b_{1}+b_{2}+\cdots+b_{n+1} \leqslant(n+1)^{\frac{3}{2}} \leqslant 4 n^{\frac{3}{2}}
\end{aligned}$$

For any positive integer $k$,
$$\begin{aligned}
\frac{1}{\frac{1}{x_{k+1}^{2}}+\frac{1}{x_{k+2}^{2}}+\cdots+\frac{1}{x_{2 k}^{2}}} & \leqslant \sqrt[k]{x_{k+1}^{2} x_{k+2}^{2} \cdots x_{2 k}^{2}} \leqslant\left(\frac{x_{k+1}+x_{k+2}+\cdots+x_{2 k}}{k}\right)^{2} \\
& <\frac{\left(4(2 k)^{\frac{3}{2}}\right)^{2}}{k^{2}}
\end{aligned}$$

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 $\sum_{k=0}^{2^{n}} \frac{1}{b_{k} b_{k+1}} \geqslant \frac{n}{2^{7}}$, which contradicts equation (3). Therefore, $a_{n}=1(n \geqslant 0)$