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 $j\ne k$.

  • 1
    $\begingroup$ 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 $a_i \geq 0$.) $\endgroup$ – Kaind Sep 24 '18 at 20:01
  • $\begingroup$ Indeed, that's a much simpler point of view. $\endgroup$ – Martin Argerami Sep 24 '18 at 20:17
  • $\begingroup$ Thanks @MartinArgerami for the detailed explanation. $\endgroup$ – learner Sep 24 '18 at 21:02

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