Example $\downarrow$ Find all $n \in \mathbf{N}^{*}$, such that the system of equations $\left\{\begin{array}{l}x_{1}+x_{2}+\cdots+x_{n}=9 \\ \frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}}=1\end{array}\right.$ has positive solutions $x_{1}, x_{2}, \cdots, x_{n}$. For the $n$ found, provide the corresponding solutions.
Solve $9 \times 1=\left(x_{1}+x_{2}+\cdots+x_{n}\right)\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}}\right) \geqslant n^{2}$, so $n \leqslant 3$.

If $n=3$, then the above inequality becomes an equality. This is only possible when $x_{1}=x_{2}=x_{3}=3$ (and it actually holds). Therefore, $n=3$ satisfies the problem conditions, and in this case, the system of equations has a unique positive integer solution $(3,3,3)$.
If $n=1$, it is obviously contradictory.
If $n=2$, then $\left\{\begin{array}{l}x_{1}+x_{2}=9 \\ \frac{1}{x_{1}}+\frac{1}{x_{2}}=1\end{array}\right.$, which is $\left\{\begin{array}{l}x_{1}+x_{2}=9 \\ x_{1} \cdot x_{2}=9\end{array}\right.$,
Solving it, we get $x_{1}=\frac{9+3 \sqrt{5}}{2}, x_{2}=\frac{9-3 \sqrt{5}}{2}$ or $x_{1}=\frac{9-3 \sqrt{5}}{2}$ and $x_{2}=\frac{9+3 \sqrt{5}}{2}$,
Since $x_{1}, x_{2}>0$,
Therefore, $n=2$ satisfies the problem conditions.
In summary, the solutions to this problem are: when $n=2$, $\left(\frac{9+3 \sqrt{5}}{2}, \frac{9-3 \sqrt{5}}{2}\right)$ or $\left(\frac{9-3 \sqrt{5}}{2}, \frac{9+3 \sqrt{5}}{2}\right)$; when $n=3$, $(3,3,3)$.
Additionally, inequality is often used as a tool for estimation in covering or embedding problems. Due to the limited space, only one example is given.