由已知条件,可知有\[\left ( 1-a_{i} \right )\left ( 1+a_{i} \right )< 1\]
\[ \therefore \left ( 1+a_{i} \right )< \frac{1}{1-a_{i}}\]
\[\prod_{i=1}^{n}\left ( 1+a_{i} \right )< \frac{1}{\prod_{i=1}^{n}\left ( 1-a_{i} \right )}\]
\[\prod_{i=1}^{n}\left ( 1-a_{i} \right )=1-\sum_{i=1}^{n}a_{i}+Y<1-\sum_{i=1}^{n}a_{i} \]
(由归纳法可证: 余项 Y>0)
\[\therefore \frac{1}{\prod_{i=1}^{n}\left ( 1-a_{i} \right )} < \frac{1}{1-\sum_{i=1}^{n}a_{i}}\]
\[\therefore \frac{1}{1-\sum_{i=1}^{n}a_{i}}> \prod_{i=1}^{n}\left ( 1+a_{i} \right )\]