Bernoulli's Inequality/Corollary/General Result

Theorem
For all $n \in \Z_{\ge 0}$:


 * $\displaystyle \prod_{j \mathop = 1}^n \left({1 - a_j}\right) \ge 1 - \sum_{j \mathop = 1}^n a^j$

where $0 < a_j < 1$ for all $j$.

Proof
The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \prod_{j \mathop = 1}^n \left({1 - a_j}\right) \ge 1 - \sum_{j \mathop = 1}^n a^j$

$P \left({0}\right)$ is the case:

Thus $P \left({0}\right)$ is seen to hold.

Basis for the Induction
$P \left({1}\right)$ is the case:

Thus $P \left({1}\right)$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is the induction hypothesis:
 * $\displaystyle \prod_{j \mathop = 1}^k \left({1 - a_j}\right) \ge 1 - \sum_{j \mathop = 1}^k a^j$

from which it is to be shown that:
 * $\displaystyle \prod_{j \mathop = 1}^{k + 1} \left({1 - a_j}\right) \ge 1 - \sum_{j \mathop = 1}^{k + 1} a^j$

Induction Step
This is the induction step:

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \forall n \in \Z_{\ge 0}: \prod_{j \mathop = 1}^n \left({1 - a_j}\right) \ge 1 - \sum_{j \mathop = 1}^n a^j$