Weierstrass product inequality

For any finite family (ai)iI of real numbers in the interval [0,1], we have


Proof: Write


For any kI, and any fixed values of the ai for ik, f is a polynomialPlanetmathPlanetmath of the first degree in ak. Consequently f is minimal either at ak=0 or ak=1. That brings us down to two cases: all the ai are zero, or at least one of them is 1. But in both cases it is clear that f1, QED.

Title Weierstrass product inequalityMathworldPlanetmath
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