Is there a 'simple' way to prove that $f\le F$ for any $\textit{n }$and for all $0
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} $