Ky Fan Inequality - Proof of The Generalization

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 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 (0,½], 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

\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)\\ &<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{align}

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

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