Maximize a series I have a set of positive numbers $X = \{x_1, x_2, x_3 \dots x_n \}$ such that $\sum x_i = m$. 
I am trying to maximize the following summation, 
$$ S = \sum_{x_i \in X, x_j\in X} (x_ix_j)^3 $$
I was wondering at what value of $x_i$ will the summation $S$ be maximized. 

By symmetry, the answer looks to me as $x_i = m/n$ but I am not sure what is the proof.
 A: Note that you have
$$
S^{1/2}=\sum_{j=1}^n x_j^3,
$$
so you only need to maximize the latter sum. If you apply Lagrange multipliers, your equations will be 
$$
3x_k^2-\lambda=0,\ k=1,\ldots,n
$$
so the only critical point occurs when $x_1=x_2=\cdots=x_n$. It is easy to see that this point has to be a minimum (because $(2t)^3+0 > t^3+ t^3$ for all $t>0$). 
Our sum will have a maximum if we have a compact domain; we achieve this by allowing $0\leq x_j$. Concretely, we  are maximizing continuous function $f(x_1,\ldots,x_n)=\sum_{j=1}^nx_j^3$ on the compact set 
$$
\{(x_1,\ldots,x_n):\ \sum_{j=1}^n x_j=m\}\cap [0,m]\times\cdots\times[0,m]. 
$$
So the maximum occurs on the boundary; that means that one or more of the $x_j$ has to be zero. The roles of the $x_j$ are symmetric so let's assume that $x_n=0$. The above argument with Lagrange Multipliers, applied now to $x_1,\ldots,x_{n-1}$ shows only a minimum in the interior, so the maximum has to be in the boundary. Repeating this argument leads us to $x_1=m$, $x_2=\cdots=x_n=0$, where the maximum is achieved. 
In summary, if you require $x_j>0$, there is no maximum. If you allow $x_j=0$, the maximum is $m^3$, achieved when $x_k=m$ for some $k$, and $x_j=0$ for $j\ne k$. 
