Let $x, y$ be positive numbers and let $s$ denote the smallest of the numbers $x, y+\frac{1}{x}, \frac{1}{y}$. What is the largest possible value of $s$, and for which numbers $x, y$ does it occur? (H)
Let $s$ be the smallest of the numbers $x, y+\frac{1}{x}, \frac{1}{y}$, hence

$$
s \leq x, \quad s \leq y+\frac{1}{x}, \quad s \leq \frac{1}{y} \quad \text { (and equality holds at least in one place). }
$$

From the reciprocals of the first and third inequalities (since $x>0, y>0$, $s$ is also positive)

$$
\frac{1}{x} \leq \frac{1}{s}, \quad y \leq \frac{1}{s}
$$

Substituting these into the second inequality

$$
s \leq y+\frac{1}{x} \leq \frac{1}{s}+\frac{1}{s}=\frac{2}{s}
$$

Thus (again, since $s>0$)

$$
s^{2} \leq 2 \quad \text { hence } \quad s \leq \sqrt{2}
$$

Let's examine whether $s$ can take this value, and if so, when. It can be seen that in the case of equality

$$
y+\frac{1}{x}=s, \quad \frac{1}{x}=\frac{1}{s}, \quad y=\frac{1}{s}
$$

thus $x=\sqrt{2}, y=\frac{1}{\sqrt{2}}$. In this case, indeed, equality holds:

$$
x=\frac{1}{y}=y+\frac{1}{x}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}
$$

Based on the work of Márton Makai (Debrecen, Fazekas M. Gymnasium, III. grade)

Note. With the notation $x_{1}=x, x_{2}=\frac{1}{y}$ ($\left(x_{1}, x_{2}>0\right)$)

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

Similarly, for any positive numbers $x_{1}, \ldots, x_{n}$

$$
s_{n}=\min \left(x_{1}, x_{2}, \ldots, x_{n}, \frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}}\right)
$$

can be formed, and it can be easily proven in the manner of the above solution that the largest possible value of $s_{n}$ is $\sqrt{n}$, and it is attained when $x_{1}=x_{2}=\cdots=x_{n}=\sqrt{n}$.

Based on the work of Balázs Németh (Budapest, Szent István Gymnasium, II. grade)