169. Let $x, y$ be positive numbers, $s$ be the smallest of the numbers $x, y+\frac{1}{x}, \frac{1}{y}$. Find the greatest possible value of $s$. For which $x$ and $y$ is it achieved?
169. Answer: the greatest possible value of $s$ is $\sqrt{2}$, which is achieved when $x=\sqrt{2}, y=\sqrt{2} / 2$.

We need to determine the greatest $s$ for which the inequalities $x \geqslant s, y+1 / x \geqslant s$, $1 / y \geqslant s$ can hold simultaneously (at least one of them must be an equality). It is clear that $s$ can be taken as positive - for example, for $s=1$ the three inequalities are consistent (for instance, the solution $x=1, y=1$ works). From these three inequalities, the following can be derived: $y \leqslant 1 / s, 1 / x \leqslant 1 / s, s \leqslant y+1 / x \leqslant 2 / s$, from which $s^{2} \leqslant 2, \quad s \leqslant \sqrt{2}$. The case where all inequalities become equalities is possible: $y=1 / x=\sqrt{2} / 2$, in which case $s=\sqrt{2}$.