1. Given the positive real numbers $x_{1}, x_{2}, x_{3}$, and $x_{4}$ such that

$$
x_{1}+x_{2}+x_{3}+x_{4}=1
$$

Let $s$ be the largest of the numbers

$$
\frac{x_{1}}{1+x_{1}}, \frac{x_{2}}{1+x_{1}+x_{2}}, \frac{x_{3}}{1+x_{1}+x_{2}+x_{3}}, \frac{x_{4}}{1+x_{1}+x_{2}+x_{3}+x_{4}}
$$

What is the smallest value that $s$ can achieve?
Solution. If $s$ is the largest of the numbers, then $\frac{1}{s}$ is the smallest among the numbers

$$
\frac{1+x_{1}}{x_{1}}, \frac{1+x_{1}+x_{2}}{x_{2}}, \frac{1+x_{1}+x_{2}+x_{3}}{x_{3}}, \frac{1+x_{1}+x_{2}+x_{3}+x_{4}}{x_{4}}
$$

Furthermore, $\frac{1}{s}-1=\frac{1-s}{s}$ is the smallest among the numbers

$$
\frac{1+x_{1}}{x_{1}}-1=\frac{1}{x_{1}}, \frac{1+x_{1}+x_{2}}{x_{2}}-1=\frac{1+x_{1}}{x_{2}}, \frac{1+x_{1}+x_{2}+x_{3}}{x_{3}}-1=\frac{1+x_{1}+x_{2}}{x_{3}}, \frac{1+x_{1}+x_{2}+x_{3}+x_{4}}{x_{4}}-1=\frac{1+x_{1}+x_{2}+x_{3}}{x_{4}}
$$

and $\frac{s}{1-s}$ is the largest among the numbers

$$
x_{1}, \frac{x_{2}}{1+x_{1}}, \frac{x_{3}}{1+x_{1}+x_{2}}, \frac{x_{4}}{1+x_{1}+x_{2}+x_{3}}
$$

Finally, $\frac{s}{1-s}+1=\frac{1}{1-s}$ is the largest among the numbers

$$
x_{1}+1, \frac{1+x_{1}+x_{2}}{1+x_{1}}, \frac{1+x_{1}+x_{2}+x_{3}}{1+x_{1}+x_{2}}, \frac{1+x_{1}+x_{2}+x_{3}+x_{4}}{1+x_{1}+x_{2}+x_{3}}
$$

Since

$$
\left(x_{1}+1\right) \cdot \frac{1+x_{1}+x_{2}}{1+x_{1}} \cdot \frac{1+x_{1}+x_{2}+x_{3}}{1+x_{1}+x_{2}} \cdot \frac{1+x_{1}+x_{2}+x_{3}+x_{4}}{1+x_{1}+x_{2}+x_{3}}=1+x_{1}+x_{2}+x_{3}+x_{4}=2 \leq\left(\frac{1}{1-s}\right)^{4}
$$

It follows that $\frac{1}{1-s} \geq \sqrt[4]{2}$, so $s \geq 1-\frac{1}{\sqrt[4]{2}}$. Therefore, the smallest value $s$ can achieve is $1-\frac{1}{\sqrt[4]{2}}$ and it is achieved for

$$
x_{1}+1=\sqrt[4]{2}, \frac{1+x_{1}+x_{2}}{1+x_{1}}=\sqrt[4]{2}, \frac{1+x_{1}+x_{2}+x_{3}}{1+x_{1}+x_{2}}=\sqrt[4]{2}, \frac{1+x_{1}+x_{2}+x_{3}+x_{4}}{1+x_{1}+x_{2}+x_{3}}=\sqrt[4]{2}
$$

From the last equalities, we get

$$
\begin{aligned}
& x_{1}=\sqrt[4]{2}-1, x_{2}=(\sqrt[4]{2}-1) \sqrt[4]{2}=\sqrt[4]{4}-\sqrt[4]{2} \\
& x_{3}=(\sqrt[4]{2}-1) \sqrt[4]{2^{2}}=\sqrt[4]{8}-\sqrt[4]{4}, x_{4}=(\sqrt[4]{2}-1) \sqrt[4]{2^{3}}=2-\sqrt[4]{8}
\end{aligned}
$$