# Weierstrass product inequality

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

 $\prod_{i}(1-a_{i})\geq 1-\sum_{i}a_{i}\;.$

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 $i\neq k$, $f$ 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 $f\geq 1$, QED.

Title Weierstrass product inequality WeierstrassProductInequality 2013-03-22 13:58:23 2013-03-22 13:58:23 Daume (40) Daume (40) 5 Daume (40) Theorem msc 26D05