$$ \forall n \in \mathbb{N}\backslash{0} \quad \forall x_1...x_n \in [0,1] $$ $$ 1-\sum_{i=1}^n{x_i} \leq \prod_{i = 1}^{n} {(1-a_{i})} $$
I'm trying to solve this one for a long time, I've tried to use induction but then I realized it can't be applied on real nummbers.
How should one solve it?
You can use induction: on $n$. First note that the inequality is an equality when $n = 1$, which is the base case.
Next suppose that the inequality holds for $n = k$. Then let $x_{k + 1} \in [0, 1]$ be arbitrary and observe that then using the induction hypothesis we have $$ 1 - \sum_{i = 1}^{k + 1} x_i = \left(1 - \sum_{i = 1}^k x_i\right) - x_{k + 1} \leq \prod_{i = 1}^k ( 1 - x_i) - x_{k + 1}. $$ But since each $x_i \in [0,1]$ then $1 - x_i \in [0,1]$, and therefore $\prod_{i = 1}^k ( 1 - x_i) \in [0, 1]$. Thus $x_{k + 1} \geq x_{k + 1}\prod_{i = 1}^k ( 1 - x_i)$ and so we also have (factoring out the entire product $\prod_{i = 1}^k ( 1 - x_i)$) $$ \prod_{i = 1}^k ( 1 - x_i) - x_{k + 1} \leq \prod_{i = 1}^k ( 1 - x_i) - x_{k + 1} \prod_{i = 1}^k ( 1 - x_i) \leq (1 - x_{k + 1}) \prod_{i = 1}^k ( 1 - x_i) = \prod_{i = 1}^{k + 1} ( 1 - x_i). $$ Putting these two inequalities together, this completes the proof.
