Weierstrass product inequality
For any finite family (ai)i∈I of real numbers in the interval [0,1], we have
∏i(1-ai)≥1-∑iai. |
Proof: Write
f=∏i(1-ai)+∑iai. |
For any k∈I, and any fixed values of the ai for i≠k,
f is a polynomial 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 f≥1, QED.
Title | Weierstrass product inequality![]() |
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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 |