Example 1. Let positive numbers $a_{1}, a_{2}, \cdots, a_{\mathrm{n}}$ satisfy $a_{1}+a_{2}+\cdots+a_{\mathrm{n}}=1$. Find
$$
\begin{array}{l}
\frac{a_{1}}{1+a_{2}+\cdots+a_{\mathrm{n}}}+\frac{a_{2}}{1+a_{1}+a_{3}+\cdots+a_{n}} \\
+\cdots+\frac{a_{1}}{1+a_{1}+\cdots+a_{\mathrm{n}-1}}
\end{array}
$$

the minimum value. (82 West Germany Competition Problem)
Let the sum of the original expression be $S$, then
$$
S=\sum_{k=1}^{n} \frac{a_{k}}{2-a_{k}}=2 \sum_{k=1}^{n} \frac{1}{2-a_{k}}-n .(1)
$$

Since $2-a_{k}>0(k=1,2 ; \cdots ; n)$, by the Cauchy-Schwarz inequality (corollary), we have
$$
\sum_{k=1}^{n} \frac{1}{2-a_{k}} \geqslant \frac{n^{2}}{\sum_{k=1}^{n}\left(2-a_{k}\right)}=\frac{n^{2}}{2 n-1} \text {. }
$$

Combining (1) and (2), we get
$$
s \geqslant \frac{2 n^{2}}{2 n-1}-n=\frac{n}{2 n-1},
$$

where equality holds if and only if (2) holds with equality, which is true if and only if $a_{1}=a_{2}=\cdots=a_{\mathrm{n}}=\frac{1}{n}$. Therefore, the minimum value of $S$ is $\frac{n}{2 n-1}$.
Example 2. Let $P$ be a point inside $\triangle A B C$, and let $D, E, F$ be the feet of the perpendiculars from $P$ to $B C, C A, A B$, respectively. Find all points $P$ for which
$$
\frac{B C}{P D}+\frac{C A}{P F}+\frac{A B}{P F}
$$

is minimized. (IMO-22-1)
The solution can be found in "Middle School Mathematics" 1987, Issue 5, p.11.