Question 1 In the Cartesian coordinate system, points $A, B, C$ lie on the hyperbola $xy=1$, satisfying that $\triangle ABC$ is an isosceles right triangle. Find the minimum value of the area of $\triangle ABC$.
(2020, National High School Mathematics Joint Competition)
By symmetry, without loss of generality, let $\angle A=90^{\circ}$. According to the problem, we can set point $A\left(a, \frac{1}{a}\right), B\left(b, \frac{1}{b}\right)$, $C\left(c, \frac{1}{c}\right) (a, b, c$ are distinct $)$.
Since $\angle A=90^{\circ} \Rightarrow \overrightarrow{A B} \cdot \overrightarrow{A C}=0$
$\Rightarrow\left(b-a, \frac{1}{b}-\frac{1}{a}\right) \cdot\left(c-a, \frac{1}{c}-\frac{1}{a}\right)=0$
$\Rightarrow(b-a)(c-a)+\left(\frac{1}{b}-\frac{1}{a}\right)\left(\frac{1}{c}-\frac{1}{a}\right)=0$
$\Rightarrow(b-a)(c-a)+\frac{a-b}{a b} \cdot \frac{a-c}{a c}=0$
$\Rightarrow(b-a)(c-a)\left(1+\frac{1}{a^{2} b c}\right)=0$.
Since $a, b, c$ are distinct, we have
$$
\begin{array}{l}
(b-a)(c-a) \neq 0 \Rightarrow 1+\frac{1}{a^{2} b c}=0 \\
\Rightarrow a^{2} b c=-1 \Rightarrow b c=-\frac{1}{a^{2}} .
\end{array}
$$

By symmetry, without loss of generality, let $b$ have the same sign as $a$, and $c$ have the opposite sign of $a$. Let $b=\frac{\alpha}{a}\left(\alpha \in \mathbf{R}_{+}, \alpha \neq a^{2}\right)$, then $c=-\frac{1}{\alpha a}$.

Thus, $A\left(a, \frac{1}{a}\right), B\left(\frac{\alpha}{a}, \frac{a}{\alpha}\right), C\left(-\frac{1}{\alpha a},-\alpha a\right)$.
Since $A B=A C \Rightarrow A B^{2}=A C^{2}$
$$
\begin{array}{l}
\Rightarrow\left(\frac{\alpha}{a}-a\right)^{2}+\left(\frac{a}{\alpha}-\frac{1}{a}\right)^{2} \\
\quad=\left(-\frac{1}{\alpha a}-a\right)^{2}+\left(-\alpha a-\frac{1}{a}\right)^{2} \\
\Rightarrow\left(a-\frac{\alpha}{a}\right)^{2}+\left(\frac{a}{\alpha}-\frac{1}{a}\right)^{2} \\
\quad=\left(a+\frac{1}{\alpha a}\right)^{2}+\left(\alpha a+\frac{1}{a}\right)^{2} \\
\Rightarrow\left(a^{2}-\frac{1}{a^{2}}\right)\left(\frac{1}{\alpha}-\alpha\right)=4 .
\end{array}
$$

Then $4 S_{\triangle A B C}=2 A B^{2}=A B^{2}+A C^{2}$
$$
\begin{array}{l}
=\left(\frac{\alpha}{a}-a\right)^{2}+\left(\frac{a}{\alpha}-\frac{1}{a}\right)^{2}+ \\
\left(\frac{-1}{\alpha a}-a\right)^{2}+\left(-\alpha a-\frac{1}{a}\right)^{2} \\
=\left(\alpha^{2}+2+\frac{1}{\alpha^{2}}\right) a^{2}+\left(\alpha^{2}+2+\frac{1}{\alpha^{2}}\right) \frac{1}{a^{2}} \\
=\left(\alpha+\frac{1}{\alpha}\right)^{2}\left(a^{2}+\frac{1}{a^{2}}\right) \\
=\left(a^{2}+\frac{1}{a^{2}}\right)\left(\left(\alpha-\frac{1}{\alpha}\right)^{2}+4\right) \\
=\left(a^{2}+\frac{1}{a^{2}}\right)\left(\left(\frac{4}{a^{2}-\frac{1}{a^{2}}}\right)^{2}+4\right) \\
=4\left(a^{2}+\frac{1}{a^{2}}\right) \frac{\left(a^{2}+\frac{1}{a^{2}}\right)^{2}}{\left(a^{2}-\frac{1}{a^{2}}\right)^{2}} \text {. } \\
\end{array}
$$

Let $\left(a^{2}-\frac{1}{a^{2}}\right)^{2}=x\left(x \in \mathbf{R}_{+}\right)$. Then
$$
\left(a^{2}+\frac{1}{a^{2}}\right)^{2}=\left(a^{2}-\frac{1}{a^{2}}\right)^{2}+4=x+4 \text {. }
$$

Thus, $S_{\triangle A B C}=\left(a^{2}+\frac{1}{a^{2}}\right) \frac{\left(a^{2}+\frac{1}{a^{2}}\right)^{2}}{\left(a^{2}-\frac{1}{a^{2}}\right)^{2}}$
$$
\begin{array}{l}
=\frac{(x+4)^{\frac{3}{2}}}{x}=\frac{\left(\frac{x}{2}+\frac{x}{2}+4\right)^{\frac{3}{2}}}{x} \\
\geqslant \frac{\left(3 \sqrt[3]{\left(\frac{x}{2}\right)^{2} \times 4}\right)^{\frac{3}{2}}}{x}=\frac{\left(3 \sqrt[3]{x^{2}}\right)^{\frac{3}{2}}}{x}=\frac{3^{\frac{3}{2}} x}{x} \\
=3^{\frac{3}{2}}=3 \sqrt{3} \text {, } \\
\end{array}
$$

When $\frac{x}{2}=4$, i.e., $a^{2}-\frac{1}{a^{2}}=2 \sqrt{2}, \alpha-\frac{1}{\alpha}=-\sqrt{2}$, or
$$
|a|=\sqrt{\sqrt{2}+\sqrt{3}}, \alpha=\frac{\sqrt{6}-\sqrt{2}}{2}
$$

the equality holds.
In summary, the minimum value sought is $3 \sqrt{3}$.