Let n be a positive integer and let a1....an in the range [0,1] be real numbers. Show $$1-\sum_{i=1}^{n}{a_{i}}\leq \prod_{j=1}^{n}\left (1-a_{j}\right)$$ using induction. I've shown P(1) to be true. Assume $$1 - \sum_{i = 1}^n a_i \leq \prod_{j = 1}^{n} (1 - a_j)$$
LHS: $$1 - \sum_{i = 1}^{n+1} a_i = 1 - \sum_{i = 1}^n a_i - a_{n+1} \leq \prod_{j = 1}^{n} (1 - a_j) - a_{n+1}$$
Then RHS: $$\begin{align} \prod_{j = 1}^{n+1} (1 - a_j) &= \left(\prod_{j = 1}^{n} (1 - a_j)\right)(1 - a_{n+1}) \\ &= \prod_{j = 1}^{n} (1 - a_j) - \left(\prod_{j = 1}^{n} (1 - a_j)\right)a_{n+1} \\ &\geq \prod_{j = 1}^{n} (1 - a_j) - a_{n+1} \text{ because } (1 - a_j) \leq 1 \Rightarrow \left(\prod_{j = 1}^{n} (1 - a_j)\right) \leq 1. \end{align}$$
Anything I can improve on? Maybe make it more specific?