Ky Fan inequality

If with 0\le x_i\le \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} } \le \frac{ \frac1n \sum_{i=1}^n x_i } { \frac1n \sum_{i=1}^n (1-x_i) } with equality if and only if x1 = x2 = ⋅ ⋅ ⋅ = xn. Remark Let :A_n:=\frac1n\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, . . ., xn, and let :A_n':=\frac1n\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 − xn. Then the Ky Fan inequality can be written as :\frac{G_n}{G_n'}\le\frac{A_n}{A_n'}, which shows the similarity to the inequality of arithmetic and geometric means given by Gn ≤ An. ==Generalization with weights==

If xi ∈ [0,] and γi ∈ [0,1] for i = 1, . . ., n are real numbers satisfying γ1 + . . . + γn = 1, then : \frac{ \prod_{i=1}^n x_i^{\gamma_i} } { \prod_{i=1}^n (1-x_i)^{\gamma_i} } \le \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 • γixi = 0 for all i = 1, . . ., n or • all xi > 0 and there exists x ∈ (0,] 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 :f(x):= \ln x-\ln(1-x) = \ln\frac x{1-x},\qquad x\in(0,\tfrac12]. 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 ith term can be omitted. Therefore, we may assume that γi > 0 for all i in the following. If x1 = x2 = . . . = xn, then equality holds. It remains to show strict inequality if not all xi are equal. The function f is strictly concave on (0,], because we have for its second derivative :f''(x)=-\frac1{x^2}+\frac1{(1-x)^2} Using the functional equation for the natural logarithm and Jensen's inequality for the strictly concave f, we obtain that : \begin{align} \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)\\ & where we used in the last step that the γi sum to one. Taking the exponential of both sides gives the Ky Fan inequality. ==References==