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Ky Fan inequality

In mathematics, the term Ky Fan inequality refers to an inequality involving the geometric mean and arithmetic mean of two sets of real numbers within the unit interval. The result was published on page 5 of the book Inequalities by Edwin F. Beckenbach and Richard E. Bellman (1961), who attribute it to an unpublished result by Ky Fan. They discuss the inequality in connection with the inequality of arithmetic and geometric means and Augustin Louis Cauchy's proof of that inequality via forward-backward induction—a method that can also be used to prove the Ky Fan inequality.

This inequality is a special case of Levinson's inequality and serves as a foundation for several generalizations and refinements, some of which are referenced below.

Statement of the classical version

If with for i = 1, ..., n, then

with equality if and only if x<sub>1</sub>&nbsp;=&nbsp;x<sub>2</sub>&nbsp;=&nbsp;⋅&nbsp;⋅&nbsp;⋅&nbsp;=&nbsp;x<sub>n</sub>.

Remark

Let

denote the arithmetic and geometric mean, respectively, of x<sub>1</sub>, .&nbsp;.&nbsp;., x<sub>n</sub>, and let

denote the arithmetic and geometric mean, respectively, of 1&nbsp;&minus;&nbsp;x<sub>1</sub>, .&nbsp;.&nbsp;., 1&nbsp;&minus;&nbsp;x<sub>n</sub>. Then the Ky Fan inequality can be written as

which shows the similarity to the inequality of arithmetic and geometric means given by G<sub>n</sub>&nbsp;≤&nbsp;A<sub>n</sub>.

Generalization with weights

If x<sub>i</sub>&nbsp;∈&nbsp;[0,] and γ<sub>i</sub>&nbsp;∈&nbsp;[0,1] for i&nbsp;= 1, .&nbsp;.&nbsp;., n are real numbers satisfying γ<sub>1</sub> + .&nbsp;.&nbsp;. + γ<sub>n</sub> = 1, then

with the convention 0<sup>0</sup> := 0. Equality holds if and only if either

  • γ<sub>i</sub>x<sub>i</sub> = 0 for all i&nbsp;= 1, .&nbsp;.&nbsp;., n or
  • all x<sub>i</sub>&nbsp;>&nbsp;0 and there exists x&nbsp;∈&nbsp;(0,] such that x&nbsp;=&nbsp;x<sub>i</sub> for all i&nbsp;= 1, .&nbsp;.&nbsp;., n with γ<sub>i</sub>&nbsp;>&nbsp;0.

The classical version corresponds to γ<sub>i</sub> = 1/n for all i&nbsp;= 1, .&nbsp;.&nbsp;., n.

Proof of the generalization

Idea: Apply Jensen's inequality to the strictly concave function

Detailed proof: (a) If at least one x<sub>i</sub> 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 γ<sub>i</sub>x<sub>i</sub> = 0 for all i&nbsp;= 1, .&nbsp;.&nbsp;., n.

(b) Assume now that all x<sub>i</sub> > 0. If there is an i with γ<sub>i</sub>&nbsp;=&nbsp;0, then the corresponding x<sub>i</sub>&nbsp;>&nbsp;0 has no effect on either side of the inequality, hence the i<sup>th</sup> term can be omitted. Therefore, we may assume that γ<sub>i</sub>&nbsp;>&nbsp;0 for all i in the following. If x<sub>1</sub> = x<sub>2</sub> = .&nbsp;.&nbsp;. = x<sub>n</sub>, then equality holds. It remains to show strict inequality if not all x<sub>i</sub> are equal.

The function f is strictly concave on (0,], because we have for its second derivative

Using the functional equation for the natural logarithm and Jensen's inequality for the strictly concave f, we obtain that

where we used in the last step that the γ<sub>i</sub> sum to one. Taking the exponential of both sides gives the Ky Fan inequality.

References

External links