Weierstrass product inequality

For any finite family ${(a_i)}_{i\in I}$ of real numbers in the interval $[0,1]$, we have

$$\prod _i(1-a_i)\ge 1-\sum _i{a}_i.$$

Proof: Write

$$f=\prod _i(1-a_i)+\sum _i{a}_i.$$

For any $k\in I$, and any fixed values of the $a_i$ for $i\ne k$, $f$ is a polynomial of the first degree in $a_k$. Consequently $f$ is minimal either at $a_k=0$ or $a_k=1$. That brings us down to two cases: all the $a_i$ are zero, or at least one of them is $1$. But in both cases it is clear that $f\ge 1$, QED.

Title	Weierstrass product inequality
Canonical name	WeierstrassProductInequality
Date of creation	2013-03-22 13:58:23
Last modified on	2013-03-22 13:58:23
Owner	Daume (40)
Last modified by	Daume (40)
Numerical id	5
Author	Daume (40)
Entry type	Theorem
Classification	msc 26D05