$$\begin{array}{l} \left(S-a_{i}\right)^{2} \leqslant(n-1)\left[\sum_{i=1}^{n} a_{i}^{2}-a_{i}^{2}\right] \\ \because \frac{a_{i}^{3}}{S-a_{i}}+\frac{a_{i}^{3}}{S-a_{i}}+\frac{\left(S-a_{i}\right)^{2}}{(n-1)^{3}} \\ \geqslant 3 \sqrt[3]{\left(\frac{a_{i}^{3}}{S-a_{i}}\right)^{2} \frac{\left(S-a_{i}\right)^{2}}{(n-1)^{3}}}=\frac{3 a_{i}^{2}}{n-1}, \\ \therefore \frac{a_{i}^{3}}{S-a_{i}} \geqslant \frac{1}{2}\left[\frac{3 a_{i}^{2}}{n-1}-\frac{\left(S-a_{i}\right)^{2}}{(n-1)^{3}}\right] \\ \geqslant \frac{1}{2}\left[\frac{3 a_{i}^{2}}{n-1}-\frac{\sum_{i=1}^{n} a_{i}^{2}-a_{i}^{2}}{(n-1)^{2}}\right] \\ =\frac{1}{2}\left[\frac{3(n-1) a_{i}^{2}-\sum_{i=1}^{n} a_{i}^{2}+a_{i}^{2}}{(n-1)^{2}}\right] \\ =\frac{1}{2}\left[\frac{(3 n-2) a_{i}^{2}-\sum_{i=1}^{n} a_{i}^{2}}{(n-1)^{2}}\right], \\ \text { then } \sum_{i=1}^{n} \frac{a_{i}^{3}}{S-a_{i}} \\ \geqslant \frac{1}{2}\left[\frac{(3 n-2) \sum_{i=1}^{n} a_{i}^{2}-n \sum_{i=1}^{n} a_{i}^{2}}{(n-1)^{2}}\right] \\ =\frac{1}{n-1} \sum_{i=1}^{n} a_{i}^{2}. \end{array}$$

Thus, equation (3) is proved.