Let $x$ and $y$ be positive numbers. What is the maximum possible value of the smallest of the numbers $x, y+\frac{1}{x}, \frac{1}{y}$?
I. solution. Let the smallest of the quantities $x, \frac{1}{y}, y+\frac{1}{x}$ be denoted by $m(x, y)$. Then

$$
\begin{aligned}
& 0 < x, y \implies m(x, y) \leqq \sqrt{2} \text{ follows.} \text{ Therefore, the smallest of the values } x, \frac{1}{y}, y+\frac{1}{x} \text{ cannot be greater than } \sqrt{2}. \text{ Equality is possible if and only if equality holds simultaneously in (1), (2), and (3), i.e., } x=\frac{1}{y}=\sqrt{2}. \text{ In this case, } y+\frac{1}{x}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2} \text{ and thus the sought value is } \sqrt{2}.

Gergely Dombi (Pécs, JPTE I. sz. Gyak. Ált. Isk., 7th grade)

II. solution. Let $u=\frac{1}{y}$. We seek the minimum of the quantities $\left(x, u, \frac{1}{x}+\frac{1}{u}\right)$. Generally, denote the minimum of the numbers $\left(\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}\right)$ by $m\left(\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}\right)$.

Then clearly

$$
m\left(x, u, \frac{1}{x}+\frac{1}{u}\right)=m\left(m(x, u) ; \frac{1}{x}+\frac{1}{y}\right)
$$

The smaller of two positive numbers cannot be greater than their known means. First, take their harmonic mean:

$$
\begin{aligned}
m(x, u) & \leqq \frac{2}{\frac{1}{x}+\frac{1}{u}}, \quad \text{ so } \\
m\left(x, u, \frac{1}{x}+\frac{1}{u}\right) & \leqq m\left(\frac{2}{\frac{1}{x}+\frac{1}{u}}, \frac{1}{x}+\frac{1}{y}\right)
\end{aligned}
$$

Then use the fact that the minimum of two positive numbers is not greater than their geometric mean; thus

$$
m\left(x, u, \frac{1}{x}+\frac{1}{u}\right) \leqq m\left(\frac{2}{\frac{1}{x}+\frac{1}{u}}, \frac{1}{x}+\frac{1}{u}\right) \leqq \sqrt{\frac{2}{\frac{1}{x}+\frac{1}{u}}\left(\frac{1}{x}+\frac{1}{u}\right)}=\sqrt{2}
$$

Therefore,

$$
m\left(x, u, \frac{1}{x}+\frac{1}{u}\right) \leqq \sqrt{2}
$$

and equality holds precisely when $x=u=\frac{1}{x}+\frac{1}{u}$, i.e., $x=u=\sqrt{2}$.

Remark. Using the reasoning of both solutions, it can be proven that if $x_{1}, x_{2}, \ldots, x_{n}$ are arbitrary positive numbers, then the minimum of the numbers $x_{1}, x_{2}, \ldots, x_{n}, \frac{1}{x_{1}}+\frac{1}{x_{2}}+\ldots+\frac{1}{x_{n}}$ cannot be greater than $\sqrt{n}$ and is exactly $\sqrt{n}$ if and only if $x_{1}=x_{2}=\ldots=x_{n}=\sqrt{n}$.

Similarly, it can be proven that the largest of the positive numbers $x_{1}, x_{2}, \ldots, x_{n}, \frac{1}{x_{1}}+\frac{1}{x_{2}}+\ldots+\frac{1}{x_{n}}$ is at least $\sqrt{n}$.