The inequality with the subtractions can be proven easily via mathematical induction. The one with the additions is proven identically. We can choose ${\displaystyle n=1}$  as the base case and see that for this value of ${\displaystyle n}$  we get

${\displaystyle 1-x_{1}\geq 1-x_{1}}$ 

which is indeed true. Assuming now that the inequality holds for all natural numbers up to ${\displaystyle n>1}$ , for ${\displaystyle n+1}$  we have:

${\displaystyle \prod _{i=1}^{n+1}(1-x_{i})\,\,=(1-x_{n+1})\prod _{i=1}^{n}(1-x_{i})}$ 

${\displaystyle \geq (1-x_{n+1})\left(1-\sum _{i=1}^{n}x_{i}\right)}$ 

${\displaystyle =1-\sum _{i=1}^{n}x_{i}-x_{n+1}+x_{n+1}\sum _{i=1}^{n}x_{i}}$ 

${\displaystyle =1-\sum _{i=1}^{n+1}x_{i}+x_{n+1}\sum _{i=1}^{n}x_{i}}$ 

${\displaystyle \geq 1-\sum _{i=1}^{n+1}x_{i}}$ 

which concludes the proof.