title: Ky Fan inequality
text: # Ky Fan inequality
  
  In mathematics, there are two different results that share the common name of the Ky Fan inequality. One is an inequality involving the geometric mean and arithmetic mean of two sets of real numbers of the unit interval. The result was published on page 5 of the book Inequalities by Edwin F. Beckenbach and Richard E. Bellman (1961), who refer to an unpublished result of Ky Fan. They mention the result in connection with the inequality of arithmetic and geometric means and Augustin Louis Cauchy's proof of this inequality by forward-backward-induction; a method which can also be used to prove the Ky Fan inequality.
  
  This 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.
  
  The second Ky Fan inequality is used in game theory to investigate the existence of an equilibrium.
  
  ## Statement of the classical version
  
  If with $0≤x_i≤$ 1/2 {\textstyle 0\leq x_{i}\leq {\frac {1}{2}}} for i = 1, ..., n, then
  
      ${\frac {{\bigl\prod _{i=1}^{n}x_{i}{\bigr )}^{1/n}}{{\bigl\prod _{i=1}^{n}(1-x_{i}){\bigr )}^{1/n}}}\leq {\frac {{\frac {1}{n}}\sum _{i=1}^{n}x_{i}}{{\frac {1}{n}}\sum _{i=1}^{n}(1-x_{i})}}$
  
  with equality if and only if $x_1$ = $x_2$ = · · · = $x_n.$
  
  ## Remark
  
  Let
  
      $A_{n}:={\frac {1}{n}}\sum _{i=1}^{n}x_{i},\qquad G_{n}={\biggl\prod _{i=1}^{n}x_{i}{\biggr )}^{1/n}$
  
  denote the arithmetic and geometric mean, respectively, of $x_1,$ . . ., $x_n,$ and let
  
      $A_{n}':={\frac {1}{n}}\sum _{i=1}^{n}(1-x_{i}),\qquad G_{n}'={\biggl\prod _{i=1}^{n}(1-x_{i}){\biggr )}^{1/n}$
  
  denote the arithmetic and geometric mean, respectively, of 1 − $x_1,$ . . ., 1 − $x_n.$ Then the Ky Fan inequality can be written as
  
      ${\frac {G_{n}}{G_{n}'}}\leq {\frac {A_{n}}{A_{n}'}},$
  
  which shows the similarity to the inequality of arithmetic and geometric means given by $G_n$ ≤ $A_n.$
  
  ## Generalization with weights
  
  If $x_i$ ∈ [0,½] and $γ_i$ ∈ [0,1] for i = 1, . . ., n are real numbers satisfying $γ_1$ + . . . + $γ_n$ = 1, then
  
      ${\frac {\prod _{i=1}^{n}x_{i}^{\gamma _{i}}}{\prod _{i=1}^{n}(1-x_{i})^{\gamma _{i}}}}\leq {\frac {\sum _{i=1}^{n}\gamma _{i}x_{i}}{\sum _{i=1}^{n}\gamma _{i}(1-x_{i})}}$
  
  with the convention $0^0$ := 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.
  
  ## Proof of the generalization
  
  Idea: Apply Jensen's inequality to the strictly concave function
  
      $f(x):=\ln x-\ln(1-x)=\ln {\frac {x}{1-x}},\qquad x\in (0,{\tfrac {1}{2}}].$
  
  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 ith term can be omitted. Therefore, we may assume that $γ_i$ > 0 for all i in the following. If $x_1$ = $x_2$ = . . . = $x_n,$ 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
  
      $f''(x)=-{\frac {1}{x^{2}}}+{\frac {1}{(1-x)^{2}}}<0,\qquad x\in (0,{\tfrac {1}{2}}).$
  
  Using the functional equation for the natural logarithm and Jensen's inequality for the strictly concave f, we obtain that
  
      ${\begin{aligned}\ln {\frac {\prod _{i=1}^{n}x_{i}^{\gamma _{i}}}{\prod _{i=1}^{n}(1-x_{i})^{\gamma _{i}}}}&=\ln \prod _{i=1}^{n}{\Bigl{\frac {x_{i}}{1-x_{i}}}{\Bigr )}^{\gamma _{i}}\\&=\sum _{i=1}^{n}\gamma _{i}f(x_{i})\\&<f{\biggl\sum _{i=1}^{n}\gamma _{i}x_{i}{\biggr )}\\&=\ln {\frac {\sum _{i=1}^{n}\gamma _{i}x_{i}}{\sum _{i=1}^{n}\gamma _{i}(1-x_{i})}},\end{aligned}}$
  
  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.
  
  ## The Ky Fan inequality in game theory
  
  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 $f(x,y)$ be a function from $S\times S$ to the real numbers that is lower semicontinuous in x, concave in y and has $f(z,z)\leq 0$ for all z in S. Then there exists $x^{*}\in S$ such that $f(x^{*},y)\leq 0$ for all $y\in S$. This Ky Fan Inequality is used to establish the existence of equilibria in various games studied in economics.
  
  