Hello! I have a set of function, such as

$f_1(x_1,x_2,x_3,x_4,x_5,x_6)=\frac{x_1}{1+x_1-x_6} + \frac{x_2}{1+x_2-x_5} + \frac{x_3}{1+x_3-x_4} + \frac{x_4}{1+x_4-x_3} + \frac{x_5}{1+x_5-x_2} + \frac{x_6}{1+x_6-x_1}$

$f_2(x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8,x_9,x_{10})= \frac{x_1+x_2+x_3}{1+x_1+x_2+x_3-x_{10}} + \frac{x_1+x_4+x_5}{1+x_1+x_4+x_5-x_9} + \frac{x_2+x_4+x_6}{1+x_2+x_4+x_6-x_8} + \frac{x_3+x_5+x_6}{1+x_3+x_5+x_6-x_7} + \frac{x_1+x_7+x_8}{1+x_1+x_7+x_8-x_6} + \frac{x_2+x_7+x_9}{1+x_2+x_7+x_9-x_5} + \frac{x_3+x_8+x_9}{1+x_3+x_8+x_9-x_4} + \frac{x_4+x_7+x_{10}}{1+x_4+x_7+x_{10}-x_3} + \frac{x_5+x_8+x_{10}}{1+x_5+x_8+x_{10}-x_2} + \frac{x_6+x_9+x_{10}}{1+x_6+x_9+x_{10}-x_1} $

$ f_3(x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8,x_9,x_{10})=\frac{x_1}{1+x_1-(x_8+x_9+x_{10})} + \frac{x_2}{1+x_2-(x_6+x_7+x_{10})} + \frac{x_3}{1+x_3-(x_5+x_7+x_9)} + \frac{x_4}{1+x_4-(x_5+x_6+x_8)} + \frac{x_5}{1+x_5-(x_3+x_4+x_{10})} + \frac{x_6}{1+x_6-(x_2+x_4+x_9)} + \frac{x_7}{1+x_7-(x_2+x_3+x_8)} + \frac{x_8}{1+x_8-(x_1+x_4+x_7)} + \frac{x_9}{1+x_9-(x_1+x_3+x_6)} + \frac{x_{10}}{1+x_{10}-(x_1+x_2+x_5)} $

with $\sum_i x_i=1$ and $x_i>=0$.

I want to show that the global maximum for every $f_i$ is at $\forall (i,j): x_i=x_j$.

[edit] Each function has several maxima; I want to prove that there is no bigger maximum than that of $x_i=x_j$. [/edit]

With what methode can I show that such functions are maximized by an equal distribution of $x_i$? Any suggestion or hint is very welcome.

Thank you!