# Weierstrass Product Inequality

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## 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:

\(\ds \prod_{i \mathop = 1}^{k + 1} \paren {1 - a_i}\) | \(=\) | \(\ds \paren {1 - a_{k + 1} } \prod_{i \mathop = 1}^k \paren {1 - a_i}\) | ||||||||||||

\(\ds \) | \(\ge\) | \(\ds \paren {1 - a_{k + 1} } \paren {1 - \sum_{i \mathop = 1}^k a_i}\) | as $1 - a_{k + 1} \ge 0$ | |||||||||||

\(\ds \) | \(=\) | \(\ds 1 - a_{k + 1} - \sum_{i \mathop = 1}^k a_i + a_{k + 1} \sum_{i \mathop = 1}^k a_i\) | ||||||||||||

\(\ds \) | \(\ge\) | \(\ds 1 - a_{k + 1} - \sum_{i \mathop = 1}^k a_i\) | as $a_i \ge 0$ | |||||||||||

\(\ds \) | \(=\) | \(\ds 1 - \sum_{i \mathop = 1}^{k + 1} a_i\) |

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

$\blacksquare$

## Also see

## Source of Name

This entry was named for Karl Theodor Wilhelm Weierstrass.

## Sources

*This article incorporates material from Weierstrass product inequality on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.*