I have the following equations:
$$\forall{i \in \{1,...,m\}},\ \ \ U_{i}=C\frac{n_{i}}{\sum_{j=1}^{m} n_{j}}-F_{i}(n_{i}, n_{-i})$$
where $C>0$.
I know that the functions $\forall{i},\ F_i$ are increasing functions of the corresponding $n_i$, and are symmetric in the relation of $i$, namely, $F_i(n_i,n_{-i}) = F_j(n_j,n_{-j})$ if $ (n_i,n_{-i}) = (n_j,n_{-j})$ (where $n_{-i}=(n_1,\ldots,n_{i-1}, n_{i+1}, \ldots,n_m)$ ).
I am trying to find a sufficient constraint on $F$ such that in the maximum point of $U_{i}$ $\forall{i,j}\ \ n_j=n_i=n$.
I have tried to examine the derivatives these functions:
\begin{align}
\frac{\partial U_i}{\partial n_i} = C\frac{(\sum{_{k=0}^m}n_k - n_{i})}{\left(\sum{_{k=0}^m}n_k\right)^2} - \frac{\partial F_i}{\partial n_i}=0\\[10pt]
\frac{\partial U_j}{\partial n_j} =C\frac{(\sum{_{k=0}^m}n_k - n_{j})}{\left(\sum{_{k=0}^m}n_k\right)^2} - \frac{\partial F_j}{\partial n_j}=0
\end{align}
I have tried to subtract the second equation from the first one, divide the equation by $(n_j-n_i)$ and use the mean value theorem, as follows:
$$\frac{\frac{\partial F_j}{\partial n_j}-\frac{\partial F_i}{\partial n_i}}{n_j - n_i} = -C\left/(\sum_{k=1}^m n_k)^{2}\right.$$
My conclusion was that the function has to have monotonically increasing derivative (convex).
I have seen a specific example of $F_i$ which are not convex (it is implicit function), but still fulfill the requirement. Can you help me find the needed condition ?
Thank you very much in advance :)
Edit:
A simple example for these function can be: $F_{i}(n_{i},n_{-i}) = n_{i}^2+\sum_{k=0, k\neq i}^{m}n_{k}$
Or even $F_{i}(n_{i},n_{-i}) = n_{i}$