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

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