The inequality with the subtractions can be proven easily via
mathematical induction. The one with the additions is proven identically. We can choose n = 1 as the base case and see that for this value of n we get : 1 -x_1 \geq 1 - x_1 which is indeed true. Assuming now that the inequality holds for all natural numbers up to n > 1, for n + 1 we have: : \prod_{i=1}^{n+1}(1-x_i)\,\, = (1-x_{n+1})\prod_{i=1}^{n}(1-x_i) :::::\geq (1-x_{n+1})\left(1 - \sum_{i=1}^nx_i\right) :::::= 1 - \sum_{i=1}^nx_i - x_{n+1} + x_{n+1}\sum_{i=1}^nx_i :::::= 1 - \sum_{i=1}^{n+1}x_i + x_{n+1}\sum_{i=1}^nx_i :::::\geq 1 - \sum_{i=1}^{n+1}x_i which concludes the proof. == References ==