Weierstrass Product Inequality

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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.

$\blacksquare$


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.