Weierstrass Product Inequality
If 0 ≤ a, b, c, d ≤ 1, prove that
(1 - a)(1 - b)(1 - c)(1 - d) + a + b + c + d ≥ 1 |
Proof
Let the function on the left side of the inequality be called
f(a) = (1 - a)K + a + M, |
where K and M are constants determined by the fixed values of b,c, and d. In fact, f(a) is a linear function of a, with a linear graph, and it is clear from the graph that it attains a minimum value at an endpoint of its range, i.e., at a = 0 or a = 1; and, within useful limits, we can deduce which it is.
Rearranging things, we have
But the inequality does not favor the variable a; the same result must also hold for b,c, and d, and we conclude that the minimum value of
Accordingly,
f(a, b, c, d) ≥ f(0, 0, 0, 0) = 1 · 1 · 1 · 1 + 0 + 0 + 0 + 0 = 1 |
as required.
Of course one of the beauties of this approach is that the number of variables is inconsequential and therefore, without further ado, all other cases follow as immediate corollaries.
References
- R. Honsberger, More Mathematical Morsels, MAA, 1991
|Contact| |Front page| |Contents| |Generalizations| |Algebra|
Copyright © 1996-2018 Alexander Bogomolny
69655914