Prove that for positive real numbers $x_{i}, y_{i}(i=1,2,3)$, we have
$$\prod\left(x_{i}^{3}+y_{i}^{3}\right) \geqslant\left(\prod x_{i}+\prod y_{i}\right)^{3}$$

In fact, by the AM-GM inequality, we get
$$\begin{array}{l} \sqrt[3]{\frac{x_{1}^{3} x_{2}^{3} x_{3}^{3}}{\prod\left(x_{i}^{3}+y_{i}^{3}\right)}} \leqslant \frac{1}{3}\left(\sum_{i=1}^{3} \frac{x_{i}^{3}}{x_{i}^{3}+y_{i}^{3}}\right) \\ \sqrt[3]{\frac{y_{1}^{3} y_{2}^{3} y_{3}^{3}}{\prod\left(x_{i}^{3}+y_{i}^{3}\right)}} \leqslant \frac{1}{3}\left(\sum_{i=1}^{3} \frac{y_{i}^{3}}{x_{i}^{3}+y_{i}^{3}}\right) \end{array}$$

Therefore,
$$\sqrt[3]{\frac{x_{1}^{3} x_{2}^{3} x_{3}^{3}}{\prod\left(x_{i}^{3}+y_{i}^{3}\right)}}+\sqrt[3]{\frac{y_{1}^{3} y_{2}^{3} y_{3}^{3}}{\prod\left(x_{i}^{3}+y_{i}^{3}\right)}} \leqslant 1$$

That is,
$$\begin{array}{c} \prod\left(x_{i}^{3}+y_{i}^{3}\right) \geqslant\left(\prod x_{i}+\prod y_{i}\right)^{3} . \\ \text { Let } x_{1}=x_{2}=a_{k}, x_{3}=a_{k+1}, a_{n+1}=a_{1}, y_{1}=y_{2}=y_{3}=1, k=1,2, \cdots, \end{array}$$
$n$, then
$$\left(a_{k}^{3}+1\right)^{2}\left(a_{k+1}^{3}+1\right) \geqslant\left(a_{k}^{2} a_{k+1}+1\right)^{3}, k=1,2, \cdots, n$$

Multiplying them together, we get
$$\prod\left(a_{i}^{3}+1\right)^{3} \geqslant \prod\left(a_{i}^{2} a_{i+1}+1\right)^{3}$$

Thus,
$$\left(a_{1}^{3}+1\right)\left(a_{2}^{3}+1\right) \cdots\left(a_{n}^{3}+1\right) \geqslant\left(a_{1}^{2} a_{2}+1\right)\left(a_{2}^{2} a_{3}+1\right) \cdots\left(a_{n}^{2} a_{1}+1\right)$$