# 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 |