Prove that by eliminating the denominator and simplifying, the original inequality is equivalent to
$$a^{6}\left(b^{3}+c^{3}\right)+b^{6}\left(c^{3}+a^{3}\right)+c^{6}\left(a^{3}+b^{3}\right) \geqslant 2 a^{2} b^{2} c^{2}\left(a^{3}+b^{3}+c^{3}\right)$$

Because
$$2 a^{2} b^{2} c^{2}\left(a^{3}+b^{3}+c^{3}\right) \leqslant a^{5}\left(b^{4}+c^{4}\right)+b^{5}\left(c^{4}+a^{4}\right)+c^{5}\left(a^{4}+b^{4}\right)$$

And
$$\begin{array}{l} a^{6}\left(b^{3}+c^{3}\right)+b^{6}\left(c^{3}+a^{3}\right)+c^{6}\left(a^{3}+b^{3}\right) \\ a^{5}\left(b^{4}+c^{4}\right)-b^{5}\left(c^{4}+a^{4}\right)-c^{5}\left(a^{4}+b^{4}\right)= \\ a^{5} b^{3}(a-b)+a^{5} c^{3}(a-c)-b^{5} a^{3}(a-b)+ \\ b^{5} c^{3}(b-c)-c^{5} a^{3}(a-c)-c^{5} b^{3}(b-c)= \\ (a-b) a^{5} b^{3}\left(a^{2}-b^{2}\right)+(a-c) a^{3} c^{3}\left(a^{2}-c^{2}\right)+(b-c) b^{3} c^{3}\left(b^{2}-c^{2}\right)= \\ a^{3} b^{3}(a-b)^{2}(a+b)+a^{3} c^{3}(a-c)^{2}(a+c)+b^{3} c^{3}(b-c)^{2}(b+c) \geqslant 0 \end{array}$$

Therefore, inequality (1) holds.