To prove the inequality, we homogenize both ends, which is equivalent to proving:
$$\frac{1}{a^{3}(b c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2(a b c)^{\frac{4}{3}}}$$

Let $a=x^{3}, b=y^{3}, c=z^{3}$, substituting into the above inequality, we get:
$$\sum_{c y c} \frac{1}{x^{9}\left(y^{3}+z^{3}\right)} \geqslant \frac{3}{2 x^{4} y^{4} z^{4}}$$

By Muirhead's inequality, we have:

That is,
$$\begin{array}{l} \left(\sum_{s y m} x^{12} y^{12}-\sum_{s y m} x^{11} y^{8} z^{5}\right)+2\left(\sum_{s y m} x^{12} y^{9} z^{3}-\sum_{s y m} x^{11} y^{8} z^{5}\right)+ \\ \left(\sum_{s y m} x^{9} y^{9} z^{6}-\sum_{s y m} x^{8} y^{8} z^{8}\right) \geqslant 0 \end{array}$$
$$\sum_{s y m} x^{12} y^{12}+2 \sum_{s y m} x^{12} y^{9} z^{3}+\sum_{s y m} x^{9} y^{9} z^{6} \geqslant 3 \sum_{s y m} x^{11} y^{8} z^{5}+6 x^{8} y^{8} z^{8}$$

Therefore, inequality (1) holds, and thus the original inequality holds.