China Mathematical Competition

Let $c = \sqrt{a^2 - b^2}$. Then $A_1(-a, 0), A_2(a, 0), F_1(-c, 0), F_2(c, 0)$.

Denote $P(x_0, y_0)$, $Q(x_1, y_1)$, $R(x_2, y_2)$, where $\frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} = 1$, $y_0 \neq 0$.

From $QA_1 \perp PA_1, QA_2 \perp PA_2$, we have
$$\overrightarrow{A_1Q} \cdot \overrightarrow{A_1P} = (x_1 + a)(x_0 + a) + y_1y_0 = 0, \quad \textcircled{1}$$
$$\overrightarrow{A_2Q} \cdot \overrightarrow{A_2P} = (x_1 - a)(x_0 - a) + y_1y_0 = 0. \qquad \textcircled{2}$$
Subtracting ① and ②, we have $2a(x_1+x_0) = 0$, i.e. $x_1 = -x_0$. Substituting it into ①, we get $-x_0^2 + a^2 + y_1y_0 = 0$, or $y_1 = \frac{x_0^2 - a^2}{y_0}$. Then $Q(-x_0, \frac{x_0^2 - a^2}{y_0})$.

From $RF_1 \perp PF_1, RF_2 \perp PF_2$, in the same way we obtain $R(-x_0, \frac{x_0^2 - c^2}{y_0})$. Therefore,
$$|QR| = \left| \frac{x_0^2 - a^2}{y_0} - \frac{x_0^2 - c^2}{y_0} \right| = \frac{b^2}{|y_0|}.$$
Since $|y_0| \in (0, b]$, then $|QR| \ge b$, where the equality holds if and only if $|y_0| = b$ (i.e., $P(0, \pm b)$). $\square$