Example 3 For any $a>0, b>0$, find $\max \left\{\min \left\{\frac{1}{a}, \frac{1}{b}, a^{2}+b^{2}\right\}\right\}$.
(2002, Beijing Middle School Mathematics Competition)
Solution 1: Let $x=\left\{\min \frac{1}{a}, \frac{1}{b}, a^{2}+b^{2}\right\}>0$.
Then $x \leqslant \frac{1}{a} \Leftrightarrow a \leqslant \frac{1}{x}$,
$x \leqslant \frac{1}{b} \Leftrightarrow b \leqslant \frac{1}{x}$,
$x \leqslant a^{2}+b^{2}$
$\leqslant\left(\frac{1}{x}\right)^{2}+\left(\frac{1}{x}\right)^{2}=\frac{2}{x^{2}}$.
Thus, $x^{3} \leqslant 2$, i.e., $x \leqslant \sqrt[3]{2}$.
Furthermore, when equations (1), (2), and (3) hold as equalities, i.e.,
$\frac{1}{a}=\frac{1}{b}=a^{2}+b^{2} \Rightarrow a=b=\frac{1}{\sqrt[3]{2}}$,
$x$ reaches its maximum value $\sqrt[3]{2}$.
Therefore, $\max \left\{\min \left\{\frac{1}{a}, \frac{1}{b}, a^{2}+b^{2}\right\}\right\}=\sqrt[3]{2}$.
Solution 2: Let $x=\min \left\{\frac{1}{a}, \frac{1}{b}, a^{2}+b^{2}\right\}$. Then $x \leqslant \frac{1}{a}, x \leqslant \frac{1}{b}, x \leqslant a^{2}+b^{2}$.
Multiplying these inequalities, we get
$$
\begin{array}{l}
x^{3} \leqslant \frac{1}{a} \cdot \frac{1}{b}\left(a^{2}+b^{2}\right) \\
\leqslant \frac{2}{a^{2}+b^{2}}\left(a^{2}+b^{2}\right)=2 .
\end{array}
$$

Thus, $x \leqslant \sqrt[3]{2}$.
When the above inequalities hold as equalities, i.e., $\frac{1}{a}=\frac{1}{b}=a^{2}+b^{2}=\sqrt[3]{2} \Rightarrow a=b=\frac{1}{\sqrt[3]{2}}$, $x$ reaches its maximum value $\sqrt[3]{2}$. Therefore, $\max \left\{\min \left\{\frac{1}{a}, \frac{1}{b}, a^{2}+b^{2}\right\}\right\}=\sqrt[3]{2}$.