172. The sum of positive numbers $x_{1}, x_{2}, \ldots, x_{n}$ is 1. Let $s$ be the largest of the numbers

$$
\frac{x_{1}}{1+x_{1}}, \frac{x_{2}}{1+x_{1}+x_{2}}, \ldots, \frac{x_{n}}{1+x_{1}+x_{2}+\ldots+x_{n}}
$$

Find the smallest possible value of $s$. For which values of $x_{1}, x_{2}, \ldots, x_{n}$ is it achieved?
172. The smallest value of $s$, equal to $1-2^{-1 / n}$, is achieved when $x_{k}=2^{\mathrm{k} / n}\left(1-2^{-1 / n}\right)$.

Let $y_{0}=1, y_{k}=1+x_{1}+\ldots+x_{k}(1 \leqslant k \leqslant n)$. Then $y_{n}=2, x_{k}=y_{k}-y_{k-1}$, if all these numbers do not exceed $s$, i.e.

$$
\frac{x_{k}}{y_{k}}=\frac{y_{k}-y_{k-1}}{y_{k}}=1-\frac{y_{k-1}}{y_{k}} \leqslant s
$$

then $1-s \leqslant y_{k-1} / y_{k}$. Multiplying these inequalities ($k=1,2, \ldots, \ldots, n$), we get: $(1-s)^{n} \leqslant y_{0} / y_{n}=1 / 2$, from which $s \geqslant 1-2^{-1 / n}$. This value is achieved when (for all $k$) $2^{-1 / n}=1-s=$ $=y_{k-1} / y_{k}$, i.e., when $y_{k}$ form a geometric progression $y_{1}=2^{1 / n}, y_{2}=2^{2 / n}, \ldots, y_{n}=2$ with common ratio $2^{1 / n}$, and $x_{k}=$ $=2^{k / n}-2^{(k-1) / n}$