Example 1 Let $x, y \in \mathbf{R}_{+}, A=\min \left\{x, \frac{y}{x^{2}+y^{2}}\right\}$. Try to find $A_{\max }$.
Solution: We discuss in two cases.
(1) If $x \geqslant \frac{y}{x^{2}+y^{2}}$, then $A=\frac{y}{x^{2}+y^{2}}$. Thus,
$$
\begin{array}{l}
A^{2}=\left(\frac{y}{x^{2}+y^{2}}\right)^{2} \leqslant \frac{y}{x^{2}+y^{2}} \cdot x \\
=\frac{x y}{x^{2}+y^{2}} \leqslant \frac{x y}{2 x y}=\frac{1}{2} .
\end{array}
$$

Hence, $A \leqslant \frac{\sqrt{2}}{2}$.
Equality holds if and only if $x=y=\frac{\sqrt{2}}{2}$.
(2) If $x \leqslant \frac{y}{x^{2}+y^{2}}$, then $A=x$. Thus,
$$
A^{2}=x^{2} \leqslant \frac{x y}{x^{2}+y^{2}} \leqslant \frac{x y}{2 x y}=\frac{1}{2} \text {. }
$$

Hence, $A \leqslant \frac{\sqrt{2}}{2}$.
Equality holds if and only if $x=y=\frac{\sqrt{2}}{2}$.
From (1) and (2), we get $A_{\max }=\frac{\sqrt{2}}{2}$.
Comment: The main approach is to use $A^{2} \leqslant \frac{y}{x^{2}+y^{2}} \cdot x$, and then apply the squeeze theorem to transform $\frac{y}{x^{2}+y^{2}} \cdot x$ into a constant.
[Strategy Two] Let $A=\max \{a, b\}$. Then $A \geqslant \frac{a+b}{2}$.