$\left[\begin{array}{l}\text { Algebraic inequalities (miscellaneous) } \\ {[\quad \text { Case analysis }}}\end{array}\right]$

$x, y>0$. Let $S$ denote the smallest of the numbers $x, 1 / y, y+1 / x$. What is the maximum value that $S$ can take?
If $x \leq \sqrt{2}$, then $S \leq \sqrt{2}$. If $y \geq \frac{1}{\sqrt{2}}$, then $S \leq y \leq \sqrt{2}$. If $x>\sqrt{2}, y<\frac{1}{\sqrt{2}}$, then $S \leq y+\frac{1}{x}<\frac{2}{\sqrt{2}}=\sqrt{2}$. Therefore, $S \leq \sqrt{2}$. Equality is achieved when $x=\frac{1}{y}=\sqrt{2}$.

## Answer

$\sqrt{2}$.