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, b, c, d) . Obviously it is a function of four variables. Suppose that all but one of the variables, say b,c, and d, are set at some fixed values and a is allowed to vary over its range from 0 to 1. With b, c, and d fixed, f is just a function of a:

  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 f(a) = (1 - K)a + K + M, where 0 ≤ K = (1 - b)(1 - c)(1 - d) ≤ 1, since 0 ≤ b, c, d ≤ 1. Thus the slope 1 - K of the graph cannot be negative. It is immaterial whether the slope is zero or positive, for in either case the minimum value of f(a) occurs at a = 0.

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 f(a, b, c, d) is given by f(0, 0, 0, 0) (perhaps not uniquely, but that doesn't matter).

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

  1. R. Honsberger, More Mathematical Morsels, MAA, 1991

|Contact| |Front page| |Contents| |Generalizations| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny

[an error occurred while processing this directive]
[an error occurred while processing this directive]