6-101. Let $x, y$ be positive numbers, and $S$ be the smallest number among $x, y+\frac{1}{x}, \frac{1}{y}$. Find the possible maximum value of $S$, and the values of $x, y$ that achieve this maximum value.
[Solution] We consider, when $S$ takes the maximum value, it can simultaneously satisfy the three inequalities
$$
\begin{array}{l}
x \geqslant S, \\
y+\frac{1}{x} \geqslant S, \\
\frac{1}{y} \geqslant S .
\end{array}
$$

And at least one of them is an equality.
Since when $x=y=1$, $S=1$, we can assume $S>0$.

From (3) we get $y \leqslant \frac{1}{S}$,
From (1) we get $\frac{1}{x} \leqslant \frac{1}{S}$,
From (2) we get $S \leqslant \frac{1}{x}+y \leqslant \frac{2}{S}$,
which implies $S^{2} \leqslant 2$,
$$
S \leqslant \sqrt{2} .
$$

On the other hand, when $x=\sqrt{2}, y=\frac{\sqrt{2}}{2}$, we have $S=\sqrt{2}$.
Therefore, the possible maximum value of $S$ is $\sqrt{2}$, and when $x=\sqrt{2}, y=\frac{\sqrt{2}}{2}$, $S$ reaches its maximum value.