- Ky Fan inequality
In
mathematics , the Ky Fan inequality is aninequality involving thegeometric mean andarithmetic mean of two sets ofreal number s of theunit interval . The result was published on page 5 of the book "Inequalities" by Beckenbach and Bellman (1961), who refer to an unpublished result ofKy Fan . They mention the result in connection with theinequality of arithmetic and geometric means andAugustin Louis Cauchy 's proof of this inequality by forward-backward-induction; a method which can also be used to prove the Ky Fan inequality.The Ky Fan inequality is a special case of
Levinson's inequality and also the starting point for several generalizations and refinements, some of them are given in the references below.tatement of the classical version
If "xi" with 0 ≤ "xi" ≤ ½ for "i" = 1, ..., "n" are real numbers, then
:
with equality if and only if "x"1 = "x"2 = . . . = "xn".
Remark
Let:
denote the arithmetic and geometric mean, respectively, of "x"1, . . ., "xn", and let
:
denote the arithmetic and geometric mean, respectively, of 1 − "x"1, . . ., 1 − "xn". Then the Ky Fan inequality can be written as
:
which shows the similarity to the
inequality of arithmetic and geometric means given by "Gn" ≤ "An".Generalization with weights
If "xi" ∈ [0,½] and "γi" ∈ [0,1] for "i" = 1, . . ., "n" are real numbers satisfying "γ"1 + . . . + "γn" = 1, then
:
with the convention 00 := 0. Equality holds if and only if either
*"γixi" = 0 for all "i" = 1, . . ., "n" or
*all "xi" > 0 and there exists "x" ∈ (0,½] such that "x" = "xi" for all "i" = 1, . . ., "n" with "γi" > 0.The classical version corresponds to "γi" = 1/"n" for all "i" = 1, . . ., "n".
Proof of the generalization
Idea: Apply
Jensen's inequality to the strictly concave function:
Detailed proof: (a) If at least one "xi" is zero, then the left-hand side of the Ky Fan inequality is zero and the inequality is proved. Equality holds if and only if the right-hand side is also zero, which is the case when "γixi" = 0 for all "i" = 1, . . ., "n".
(b) Assume now that all "xi" > 0. If there is an "i" with "γi" = 0, then the corresponding "xi" > 0 has no effect on either side of the inequality, hence the "i"th term can be omitted. Therefore, we may assume that "γi" > 0 for all "i" in the following. If "x"1 = "x"2 = . . . = "xn", then equality holds. It remains to show strict inequality if not all "xi" are equal.
The function "f" is strictly concave on (0,½] , because we have for its second derivative
:
Using the
functional equation for thenatural logarithm and Jensen's inequality for the strictly concave "f", we obtain that:
Wikimedia Foundation. 2010.