# Weierstrass Product Inequality

## Theorem

For $n \ge 1$:

$\displaystyle \prod_{i \mathop = 1}^n \left({1 - a_i}\right) \ge 1 - \sum_{i \mathop = 1}^n a_i$

where all of $a_i$ are in the closed interval $\left[{0 \,.\,.\,1 }\right]$.

## 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:

$\displaystyle \prod_{i \mathop = 1}^k \left({1 - a_i}\right) \ge 1 - \sum_{i \mathop = 1}^k a_i$

Then:

 $\displaystyle \prod_{i \mathop = 1}^{k + 1} \left({1 - a_i}\right)$ $=$ $\displaystyle \left({1 - a_{k + 1} }\right) \prod_{i \mathop = 1}^k \left({1 - a_i}\right)$ $\displaystyle$ $\ge$ $\displaystyle \left({1 - a_{k + 1} }\right) \left({1 - \sum_{i \mathop = 1}^k a_i}\right)$ as $1 - a_{k + 1} \ge 0$ $\displaystyle$ $=$ $\displaystyle 1 - a_{k + 1} - \sum_{i \mathop = 1}^k a_i + a_{k + 1} \sum_{i \mathop = 1}^k a_i$ $\displaystyle$ $\ge$ $\displaystyle 1 - a_{k + 1} - \sum_{i \mathop = 1}^k a_i$ as $a_i \ge 0$ $\displaystyle$ $=$ $\displaystyle 1 - \sum_{i \mathop = 1}^{k + 1} a_i$

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

$\blacksquare$

## Source of Name

This entry was named for Karl Theodor Wilhelm Weierstrass.