30. As shown in the figure, draw $P X \perp B C, P Y \perp C A$, $P Z \perp A B$, with $X, Y, Z$ being the feet of the perpendiculars. Extend $X P$, and draw perpendiculars from $Y, Z$ to the extended line, with the feet of the perpendiculars being $M, N$, respectively. Since $P Y \perp C A, P Z \perp A B$, points $A, Y, P, Z$ are concyclic, and the diameter of the circle is $P A$. According to the sine rule, $Y Z = P A \sin \angle Y A Z = P A \sin A$. Since $P X \perp B C, P Y \perp C A$, we have $\angle Z P N = \angle B, \angle Y P M = \angle C$. Considering the horizontal projection of $Y Z$, we have $Y Z \geqslant Z M + N Y$, with equality holding if and only if $Y Z \perp P X$. And $Z M = P Z \sin \angle Z P N = P Z \sin B, N Y = P Y \sin \angle Y P M = P Y \sin C$. Thus, from $Y Z \geqslant Z M + N Y$, we get
$$P A \sin A \geqslant P Z \sin B + P Y \sin C$$

Multiplying both sides by $2 R$, we get
$$P A \cdot A B \geqslant P Z \cdot C A + P Y \cdot A B$$

Equality holds in inequality (1) if and only if $Y Z \perp P X$. Similarly, we can prove that
$$P B \cdot C A \geqslant P X \cdot A B + P Z \cdot B C$$

Equality holds in inequality (2) if and only if $X Z \perp P Y$.
$$P C \cdot A B \geqslant P X \cdot C A + P Y \cdot B C$$

Equality holds in inequality (3) if and only if $P Z \perp X Y$. Applying inequalities (1), (2), and (3) to the left side of the inequality and using the AM-GM inequality, we get
$$\begin{array}{l} \frac{P A}{B C^{2}} + \frac{P B}{C A^{2}} + \frac{P C}{A B^{2}} = \frac{1}{B C^{3}} \cdot P A \cdot B C + \frac{1}{C A^{3}} \cdot P B \cdot C A + \frac{1}{A B^{3}} \cdot P C \cdot A B \geqslant \\ \frac{1}{B C^{3}}(P Z \cdot C A + P Y \cdot A B) + \\ \frac{1}{C A^{3}} \cdot (P X \cdot A B + P Z \cdot B C) + \\ \frac{1}{A B^{3}} \cdot (P X \cdot C A + P Y \cdot B C) = \\ \left(\frac{A B}{C A^{3}} + \frac{C A}{A B^{3}}\right) P X + \left(\frac{A B}{B C^{3}} + \frac{B C}{A B^{3}}\right) P Y + \left(\frac{B C}{C A^{3}} + \frac{C A}{B C^{3}}\right) P Z \geqslant \\ \frac{2}{A B \cdot C A} \cdot P X + \frac{2}{A B \cdot B C} \cdot P Y + \frac{2}{B C \cdot C A} \cdot P Z = \\ \frac{4}{A B \cdot B C \cdot C A} \cdot \frac{1}{2}(B C \cdot P X + C A \cdot P Y + A B \cdot P Z) = \\ \frac{4 S_{\triangle A B C}}{A B \cdot B C \cdot C A} = \frac{1}{R} \end{array}$$