11. Given the function $y=f(x)$ satisfies $\boldsymbol{a}=\left(x^{2}, y\right), \boldsymbol{b}=\left(x-\frac{1}{x},-1\right)$, and $\boldsymbol{a} \cdot \boldsymbol{b}=-1$. If there exists a positive sequence $\left\{a_{n}\right\}$ such that: $a_{1}=\frac{1}{2}, \sum_{i=1}^{n} f\left(a_{i}\right)-n=\sum_{i=1}^{n} a_{i}^{3}-n^{2} a_{n}\left(n \in \mathbf{N}^{*}\right)$.
(1) Find the general term of the sequence $\left\{a_{n}\right\}$;
(2) Prove: $\sum_{i=1}^{n} \sqrt{\left[(i+1) \cdot a_{i}\right]^{3}}<3$.
11. Solution: (1) From the given, $x^{2}\left(x-\frac{1}{x}\right)-y=-1$, then $y=f(x)=x^{3}-x+1$, thus $\sum_{i=1}^{n}\left(a_{i}^{3}-a_{i}+1\right)-n=\sum_{i=1}^{n} a_{i}^{3}-n^{2} a_{n}$, let $S_{n}=\sum_{i=1}^{n} a_{i}$, we get $S_{n}=n^{2} a_{n}$, hence $S_{n-1}=(n-1)^{2} a_{n-1}$, subtracting the two equations gives $\frac{a_{n}}{a_{n-1}}=\frac{n-1}{n+1}(n \geqslant 2)$, thus $a_{n}=a_{1} \cdot \frac{a_{2}}{a_{1}} \cdots \frac{a_{n}}{a_{n-1}}=\frac{1}{n(n+1)}(n \geqslant 2)$, and $a_{1}=\frac{1}{2}$ also fits the above formula, so $a_{n}=\frac{1}{n(n+1)}$. (2) From (1), we get $(i+1) a_{i}=\frac{1}{i}$, thus $\sum_{i=1}^{n} \sqrt{\left[(i+1) \cdot a_{i}\right]^{3}}=\sum_{i=1}^{n} \frac{1}{\sqrt{i^{3}}}$,
$$
\text { Also, } \frac{1}{\sqrt{i^{3}}}=\frac{1}{i \sqrt{i}}=\frac{2}{i \sqrt{i}+i \sqrt{i}}<\frac{2}{i \sqrt{i-1}+(i-1) \sqrt{i}} \text {, }
$$

Therefore, when $i \geqslant 2$,
$$
\begin{array}{l}
\frac{2}{i \sqrt{i-1}+(i-1) \sqrt{i}}=\frac{2}{\sqrt{i} \sqrt{i-1}(\sqrt{i}+\sqrt{i-1})}=\frac{2(\sqrt{i}-\sqrt{i-1})}{\sqrt{i} \sqrt{i-1}}=2\left(\frac{1}{\sqrt{i-1}}-\frac{1}{\sqrt{i}}\right), \\
\text { thus } \sum_{i=1}^{n} \sqrt{\left[(i+1) \cdot a_{i}\right]^{3}}=1+\frac{1}{\sqrt{2^{3}}}+\cdots+\frac{1}{\sqrt{n^{3}}} \\
<1+2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right) \\
=3-\frac{2}{\sqrt{n}}<3 .
\end{array}
$$