# 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.

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 $$j\ne k$$.
• I was trying to solve this question using some known inequalities. I couldn't figure out which one to use then, but on seeing the answer, I realize that the inequality: $\sum_i a_i^p \leq (\sum_i a_i )^p$, ($p \in \mathbb{N}$) works and equality holds for $p>1$ iff atleast n-1 terms are zero (if $a_i \geq 0$.) – Kaind Sep 24 '18 at 20:01