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Is there a 'simple' way to prove that $f\le F$ for any $\textit{n }$and for all $0<x_{i} ,y_{i} <1$?

I have only verified this numerically...

$$ \begin{array}{l} f(x_{1} ,...,x_{n} ,y_{1} ,...,y_{n} )= \\ {\frac{1}{n^{2} } \sum _{i=1}^{n}\left(2x_{i} (1-x_{i} )+2y_{i} (1-y_{i} )-4(-x_{i} ^{4} -y_{i} ^{4} +2x_{i} ^{3} +2y_{i} ^{3} -2x_{i} ^{2} -2y_{i} ^{2} +x_{i} +y_{i} )^{2} \right) } \end{array}$$

$$F=\frac{1}{n} \left(2x(1-x)+2y(1-y)-4(-x^{4} -y^{4} +2x^{3} +2y^{3} -2x^{2} -2y^{2} +x+y)^{2} \right) $$ where, $x=\frac{1}{n} \sum _{i=1}^{n}x_{i} ,\quad y=\frac{1}{n} \sum _{i=1}^{n}y_{i} $

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I think you should try using Jensen's inequality, which says that if $g$ is a real convex function, then we have :

$g\left(\frac{1}{n}\sum x_i\right) \leq \frac{1}{n} \sum g(x_i)$

(and the inequality is reversed when $g$ is concave)

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  • $\begingroup$ Yes, I considered using Jensen's but g is a multivariate and I am not sure [a] how to use a multivariate version of Jensen's, and [b] how to determine if g is convex or concave... $\endgroup$
    – Omri
    Feb 14 '12 at 15:55

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