6・100 Let $n$ positive numbers $x_{1}, x_{2}, \cdots, x_{n}$ sum up to 1. Let $S$ be the largest of the following numbers:
$$
\frac{x_{1}}{1+x_{1}}, \frac{x_{2}}{1+x_{1}+x_{2}}, \cdots, \frac{x_{n}}{1+x_{1}+x_{2}+\cdots+x_{n}} .
$$

Find the possible minimum value of $S$. For what values of $x_{1}, x_{2}, \cdots, x_{n}$ is this minimum value achieved?
【Solution】Let $y_{0}=1, y_{k}=1+x_{1}+x_{2}+\cdots+x_{k}(1 \leqslant k \leqslant n)$.
Then
$$
x_{k}=y_{k}-y_{k-1} \text {. }
$$

From the given condition,
$$
\frac{x_{k}}{1+x_{1}+x_{2}+\cdots+x_{k}} \leqslant S,
$$

which is $\frac{y_{k}-y_{k-1}}{y_{k}} \leqslant S$,
thus we have $\quad 1-S \leqslant \frac{y_{k-1}}{y_{k}} \quad(k=1,2, \cdots, n)$
From the given, we know $S \leqslant 1$ and $y_{k}>0(k=0,1,2, \cdots, n)$, so we can multiply the $n$ inequalities expressed by (1), thus obtaining
$$
(1-S)^{n} \leqslant \frac{y_{0}}{y_{n}}
$$

Noting that $y_{0}=1, y_{n}=2$, we have
$$
\begin{array}{l}
(1-S)^{n} \leqslant \frac{1}{2}, \\
S \geqslant 1-2^{-\frac{1}{n}} .
\end{array}
$$

On the other hand, if we take
$$
\frac{y_{k-1}}{y_{k}}=1-S=2^{-\frac{1}{n}} \quad(k=1,2, \cdots, n),
$$

that is, we take
$$
y_{0}=1, y_{1}=2^{\frac{1}{n}}, y_{2}=2^{\frac{2}{n}}, y_{3}=2^{\frac{3}{n}}, \cdots, y_{n}=2 \text {, }
$$

then
$$
S=1-2^{-\frac{1}{n}} \text {. }
$$

In this case,
$$
x_{k}=y_{k}-y_{k-1}=2^{\frac{k}{n}}-2^{\frac{k-1}{n}} \quad(k=1,2, \cdots, n) .
$$

This means that when $x_{k}=2^{\frac{k}{n}}-2^{\frac{k-1}{n}}(k=1,2, \cdots, n)$, $S$ can achieve the minimum value $1-2^{-\frac{1}{n}}$.