# Global maximum of non-concave function

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 $\forall (i,j): x_i=x_j$.

 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!

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There are several possible maxima it seems. Consider, e.g., for $f_1$, that $x_1=x_6=1/2$ and all other $x$s are zero. Then, $f_1=1$; same value is also achieved by your choice of $x_i=1/6$. Or are you asking whether for an optimum we can always pick a uniform vector? This might follows from the cyclic symmetry of your function, or the other kinds of symmetry groups that these functions are encoding. –  Suvrit Apr 22 '13 at 23:26
Thanks for the answere. Numerically I also found several maxima, too. As you suggest, I want to prove that non of them can be bigger than $f(x_i)$ with $x_i=x_j$. You're right, the cyclic symmetry is notable. Do you have any hint on how to exploit this characteristic? –  Lithops Apr 22 '13 at 23:56