Problem 1476

9. (1) Using the mean inequality, we have
$$\begin{array}{l} \quad \frac{a^{5}}{b^{2}-1}+\frac{25(5+\sqrt{15})}{12}(b-1)+\frac{25(5-\sqrt{15})}{12}(b+1)+\frac{25}{18} \sqrt{15}+ \\ \frac{25}{18} \sqrt{15} \geqslant \frac{125}{6} a \text {, hence } \frac{a^{5}}{b^{2}-1} \geqslant \frac{125}{6}(a-b)+\frac{25}{18} \sqrt{15} . \end{array}$$

Similarly, $\frac{b^{5}}{c^{2}-1} \geqslant \frac{125}{6}(b-c)+\frac{25}{18} \sqrt{15} ; \frac{c^{5}}{a^{2}-1} \geqslant \frac{125}{6}(c-a)+\frac{25}{18} \sqrt{15}$. Adding these inequalities yields the original inequality.

(2) Since $\left(b^{3}-1\right)\left(b^{3}-1\right)\left(b^{3}-1\right) \times \frac{3}{2} \times \frac{3}{2} \leqslant \frac{3^{5} \cdot b^{15}}{5^{5}}$, we have $b^{3}-1 \leqslant$ $\frac{3 \cdot \sqrt[3]{20} \cdot b^{5}}{25}$, thus $\frac{a^{5}}{b^{3}-1} \geqslant \frac{5 \cdot \sqrt[3]{50} \cdot a^{5}}{6 b^{5}}$. Similarly, $\frac{b^{5}}{c^{3}-1} \geqslant \frac{5 \cdot \sqrt[3]{50} \cdot b^{5}}{6 c^{5}}$; $\frac{c^{5}}{a^{3}-1} \geqslant \frac{5 \cdot \sqrt[3]{50} \cdot c^{5}}{6 a^{5}}$. Therefore, the left side of the inequality $\geqslant \frac{5 \sqrt[3]{50}}{6}\left(\frac{a^{5}}{b^{5}}+\frac{b^{5}}{c^{5}}+\frac{c^{5}}{a^{5}}\right) \geqslant \frac{5}{2} \cdot \sqrt[3]{50}$.

Note: The first part (1) can also be solved using a method similar to part (2).
Since $\left(b^{2}-1\right)\left(b^{2}-1\right) \cdot \frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} \leqslant\left(\frac{2 b^{2}}{5}\right)^{5}$, we have $b^{2}-1 \leqslant \frac{6 \sqrt{3} b^{5}}{25 \sqrt{5}}$, thus $\frac{a^{5}}{b^{2}-1} \geqslant \frac{25 \sqrt{5} a^{5}}{6 \sqrt{3} b^{5}}$. This makes it easy to prove the original inequality.