China Mathematical Competition

When $x > 0$, $y = \frac{1}{x}$. When $x < 0$, $y = -\frac{1}{x}$, and its corresponding derivative is $y' = -\frac{1}{x^2}$.

Suppose $Q(-a, \frac{1}{a})$, where $a > 0$. By the condition, the slope of $PQ$ is $y'|_{x=-a} = \frac{1}{a^2}$.

The equation of line $PQ$ is $y = \frac{1}{a^2}(x + a) + \frac{1}{a} = \frac{x + 2a}{a^2}$.

Combining the above equation with $y = \frac{1}{x}$ ($x > 0$) yields $x^2 + 2a x - a^2 = 0$, and thus we know the abscissa $x_P = (\sqrt{2} - 1)a$ of point $P$, with the negative root discarded. Therefore,

$$\begin{align*} |PQ| &= \sqrt{1 + \left(\frac{1}{a^2}\right)^2} \cdot |x_P - x_Q| \\ &= \sqrt{1 + \frac{1}{a^4}} \cdot (\sqrt{2}a) \\ &\geq \sqrt{2\sqrt{1 \cdot \frac{1}{a^4}}} \cdot \sqrt{2}a = 2. \end{align*}$$

When $a = 1$, namely, $Q(-1, 1)$, the minimum of $|PQ|$ is $2$. $\square$