3. For $n$ positive numbers $x_{1}, x_{2}, \cdots, x_{n}$ whose sum equals 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 minimum possible value of $S$, and for what values of $x_{1}, x_{2}, \cdots, x_{n}$ is this minimum value achieved?
（6th All-Soviet Union Olympiad problem）
3. Let $y_{0}=1, y_{k}=1+x_{1}+x_{2}+\cdots+x_{k}(1 \leqslant k \leqslant n)$, then $x_{k}=y_{k}-y_{k-1}$. From the given, $\frac{x_{k}}{1+x_{1}+x_{2}+\cdots+x_{k}} \leqslant S$, which means $\frac{y_{k}-y_{k-1}}{y_{k}} \leqslant S$, thus we get $1-S \leqslant \frac{y_{k-1}}{y_{k}}(k=1,2, \cdots, n)$.

From the given, we have $S \leqslant 1$ 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 $(1-S)^{n} \leqslant \frac{1}{2}$, which means $S \geqslant 1-2^{-\frac{1}{n}}$.

On the other hand, if we take $\frac{y_{k-1}}{y_{k}}=1-S=2^{-\frac{1}{n}}(k=1,2, \cdots, n)$, i.e., take $y_{0}=1, y_{1}=2^{\frac{1}{n}}, y_{2}=2^{\frac{2}{n}}, y_{3}=$ $2^{\frac{3}{n}}, \cdots, y_{n}=2$, then $S=1-2^{-\frac{1}{n}}$. At this time, $x_{k}=y_{k}-y_{k-1}=2^{\frac{k}{n}}-2^{\frac{k-1}{n}}(k=1,2, \cdots, n)$. This shows that when $x_{k}=$ $2^{\frac{k}{n}}-2^{\frac{k-1}{n}}(k=1,2, \cdots, n)$, $S$ can achieve the minimum value $1-2^{-\frac{1}{n}}$.