Let $x_{1}, x_{2}, \ldots, x_{n}$ be positive numbers whose sum is 1. Let

$$
s=\max \left(\frac{x_{1}}{1+x_{1}}, \frac{x_{2}}{1+x_{1}+x_{2}}, \cdots, \frac{x_{n}}{1+x_{1}+\cdots+x_{n}}\right)
$$

What is the smallest possible value of $s$? For which $x_{1}, x_{2}, \ldots, x_{n}$ does it attain this value?
Let $a_{0}=1, a_{1}=1+x_{1}, a_{2}=1+x_{1}+x_{2}, \ldots, a_{n}=1+x_{1}+\ldots+x_{n}=2$. Thus,

$$
\begin{aligned}
s & =\max \left(\frac{a_{1}-a_{0}}{a_{1}}, \frac{a_{2}-a_{1}}{a_{2}}, \ldots, \frac{a_{n}-a_{n-1}}{a_{n}}\right)=\max \left(1-\frac{a_{0}}{a_{1}}, \ldots, 1-\frac{a_{n-1}}{a_{n}}\right)= \\
& =1-\min \left(\frac{a_{0}}{a_{1}}, \ldots, \frac{a_{n-1}}{a_{n}}\right) .
\end{aligned}
$$

Since we are dealing with positive numbers, the minimum of these numbers is not greater than their geometric mean, that is,

$$
s \geq 1-\sqrt[n]{\frac{a_{0}}{a_{1}} \cdot \frac{a_{1}}{a_{2}} \cdot \cdots \cdot \frac{a_{n-1}}{a_{n}}}=1-\sqrt[n]{\frac{a_{0}}{a_{n}}}=1-\sqrt[n]{\frac{1}{2}}
$$

Thus,

$$
s \geq 1-\sqrt[n]{\frac{1}{2}}
$$

This value, however, is also attained precisely when

$$
\frac{a_{0}}{a_{1}}=\frac{a_{1}}{a_{2}}=\ldots=\frac{a_{n-1}}{a_{n}}
$$

which means that the $a_{i}$ numbers are consecutive terms of a geometric sequence. Given the choice $a_{0}=1, a_{n}=2$, this implies that

$$
a_{0}=1, \quad a_{1}=\sqrt[n]{2}, \quad a_{2}=\sqrt[n]{2^{2}}, \quad \ldots, \quad a_{n}=2
$$

from which

$$
x_{1}=\sqrt[n]{2}-1, x_{2}=\sqrt[n]{2^{2}}-\sqrt[n]{2}, \ldots, x_{k}=\sqrt[n]{2^{k}}-\sqrt[n]{2^{k-1}}, \ldots, x_{n}=2-\sqrt[n]{2^{n-1}}
$$

This answers both questions of the problem.

Based on the work of Koblinger Egmont (Fóv. Fazekas M. Gyak. Gimn. II. o. t.)