Weierstrass Product Inequality

Theorem
For $n \ge 1$:


 * $\ds \prod_{i \mathop = 1}^n \paren {1 - a_i} \ge 1 - \sum_{i \mathop = 1}^n a_i$

where all of $a_i$ are in the closed interval $\closedint 0 1$.

Proof
For $n = 1$ we have:


 * $1 - a_1 \ge 1 - a_1$

which is clearly true.

Suppose the proposition is true for $n = k$, that is:


 * $\ds \prod_{i \mathop = 1}^k \paren {1 - a_i} \ge 1 - \sum_{i \mathop = 1}^k a_i$

Then:

Thus, by Principle of Mathematical Induction, the proof is complete.

Also see

 * Bounds for Finite Product of Real Numbers