Ky Fan inequality

If with ${\textstyle 0\leq x_{i}\leq {\frac {1}{2}}}$ for i = 1, ..., n, then

${\displaystyle {\frac {{\bigl (}\prod _{i=1}^{n}x_{i}{\bigr )}^{1/n}}{{\bigl (}\prod _{i=1}^{n}(1-x_{i}){\bigr )}^{1/n}}}\leq {\frac {{\frac {1}{n}}\sum _{i=1}^{n}x_{i}}{{\frac {1}{n}}\sum _{i=1}^{n}(1-x_{i})}}}$

with equality if and only if x₁ = x₂ = · · · = x_n.

Let

${\displaystyle A_{n}:={\frac {1}{n}}\sum _{i=1}^{n}x_{i},\qquad G_{n}={\biggl (}\prod _{i=1}^{n}x_{i}{\biggr )}^{1/n}}$

denote the arithmetic and geometric mean, respectively, of x₁, . . ., x_n, and let

${\displaystyle A_{n}':={\frac {1}{n}}\sum _{i=1}^{n}(1-x_{i}),\qquad G_{n}'={\biggl (}\prod _{i=1}^{n}(1-x_{i}){\biggr )}^{1/n}}$

denote the arithmetic and geometric mean, respectively, of 1 − x₁, . . ., 1 − x_n. Then the Ky Fan inequality can be written as

${\displaystyle {\frac {G_{n}}{G_{n}'}}\leq {\frac {A_{n}}{A_{n}'}},}$

which shows the similarity to the inequality of arithmetic and geometric means given by G_n ≤ A_n.

If x_i ∈ [0,½] and γ_i ∈ [0,1] for i = 1, . . ., n are real numbers satisfying γ₁ + . . . + γ_n = 1, then

${\displaystyle {\frac {\prod _{i=1}^{n}x_{i}^{\gamma _{i}}}{\prod _{i=1}^{n}(1-x_{i})^{\gamma _{i}}}}\leq {\frac {\sum _{i=1}^{n}\gamma _{i}x_{i}}{\sum _{i=1}^{n}\gamma _{i}(1-x_{i})}}}$

with the convention 0⁰ := 0. Equality holds if and only if either

• γ_ix_i = 0 for all i = 1, . . ., n or
• all x_i > 0 and there exists x ∈ (0,½] such that x = x_i for all i = 1, . . ., n with γ_i > 0.

The classical version corresponds to γ_i = 1/n for all i = 1, . . ., n.

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

${\displaystyle f(x):=\ln x-\ln(1-x)=\ln {\frac {x}{1-x}},\qquad x\in (0,{\tfrac {1}{2}}].}$

Detailed proof: (a) If at least one x_i 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_ix_i = 0 for all i = 1, . . ., n.

(b) Assume now that all x_i > 0. If there is an i with γ_i = 0, then the corresponding x_i > 0 has no effect on either side of the inequality, hence the i^th term can be omitted. Therefore, we may assume that γ_i > 0 for all i in the following. If x₁ = x₂ = . . . = x_n, then equality holds. It remains to show strict inequality if not all x_i are equal.

The function f is strictly concave on (0,½], because we have for its second derivative

${\displaystyle f''(x)=-{\frac {1}{x^{2}}}+{\frac {1}{(1-x)^{2}}}<0,\qquad x\in (0,{\tfrac {1}{2}}).}$

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

${\displaystyle {\begin{aligned}\ln {\frac {\prod _{i=1}^{n}x_{i}^{\gamma _{i}}}{\prod _{i=1}^{n}(1-x_{i})^{\gamma _{i}}}}&=\ln \prod _{i=1}^{n}{\Bigl (}{\frac {x_{i}}{1-x_{i}}}{\Bigr )}^{\gamma _{i}}\\&=\sum _{i=1}^{n}\gamma _{i}f(x_{i})\\&<f{\biggl (}\sum _{i=1}^{n}\gamma _{i}x_{i}{\biggr )}\\&=\ln {\frac {\sum _{i=1}^{n}\gamma _{i}x_{i}}{\sum _{i=1}^{n}\gamma _{i}(1-x_{i})}},\end{aligned}}}$

where we used in the last step that the γ_i sum to one. Taking the exponential of both sides gives the Ky Fan inequality.

A second inequality is also called the Ky Fan Inequality, because of a 1972 paper, "A minimax inequality and its applications". This second inequality is equivalent to the Brouwer Fixed Point Theorem, but is often more convenient. Let S be a compact convex subset of a finite-dimensional vector space V, and let ${\displaystyle f(x,y)}$ be a function from ${\displaystyle S\times S}$ to the real numbers that is lower semicontinuous in x, concave in y and has ${\displaystyle f(z,z)\leq 0}$ for all z in S. Then there exists ${\displaystyle x^{*}\in S}$ such that ${\displaystyle f(x^{*},y)\leq 0}$ for all ${\displaystyle y\in S}$. This Ky Fan Inequality is used to establish the existence of equilibria in various games studied in economics.

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• Ky Fan at the Mathematics Genealogy Project