Let $x_1,x_2..x_n$ be real numbers in $[-1,1]$ with $\sum_{i=1}^{n} {x_i}^3=0$ then find maximum value of :$A=\sum_{i=1}^n x_1$

Question

Let $x_1,x_2..x_n$ be real numbers in $[-1,1]$ with $$\sum_{i=1}^{n} {x_i}^3=0$$ then find maximum value of :$$A=\sum_{i=1}^n x_1$$

---

As i don't know how to start i give a brief background of the problem along with some of my intuition.

Background we define for each $i$: $b_i={x_i}^3$. Hence we require the minimum of: $\sum {(b_i)}^{1/3}$.The most interesting part of the problem is that $f(x)={x}^{1/3}$ is convex for all $x<0$ and concave $x>0$.

If we assume $b_1,b_2,..b_k$ as negative then we can apply the following lemma with $a=-1,b=0$.

Lemma:Suppose that f(x) is a real convex function defined on $[a,b]$ and $x_1, x_2, ... , x_n$ belonging to $[a, b]$ such that $ x_1 + x_2+ ... + x_n = S$,then the function $$F=f(x_1)+f(x_2)..+f(x_n) $$ attains maximum value if and only if $n-1$ elements are equal to $a$ or $b$.

But i am stuck on how to handle $b_{k+1}...b_n$ Jensen would work as by assumption $b_{k+1},b_{k+2}..b_n $ will be postive.But again i don't know how to start.

Answer 1

By using $\cos3\alpha=4\cos^3\alpha-3\cos\alpha$ we obtain: $$\sum_{i=1}^nx_i\leq\frac{n}{3}.$$ I hope it will help.

Answer 2

Move negative $x_i$ to the rhs, rename as $y_j$, we have $\sum_{i=1}^m x_i^3 = \sum_{j=1}^{n-m} y_j^3$ with $x_i, y_j \ge 0$, need to find max of $A = \sum_{i=1}^m x_i-\sum_{j=1}^{n-m} y_j$.

If $A$ achieve max, all $x_i$ must be equal.

If there exists some $ij$ such that $x_i < y_j < 1$, we can increase $x_i,y_j$ by a little and keep $x_i^3-y_j^3$ invariant, this will make $A$ larger. So we assume no $x_i < y_j < 1$. For similar reason, we can assume no $x_i > y_j > 0$.

In case $x_i = y_j$, replace $x_i,y_j$ with $0$, $A$ is not changed. Then $y_j$ is either $0$ or $1$. Let $k$ be the number of $y_j$'s equal to $1$ , then $A=m\cdot (\frac{k}{m})^{\frac{1}{3}}-k$ with integer $m,k$ satisfy $m+k\le n$ and $m \ge k$. $A$ increase if $m$ increases, so $m=n-k$.

Let $t =(\frac{k}{n-k})^{\frac{1}{3}}$, after some calculation $A=n(\frac{1}{t^2-t+1}-1)$, achieve max when $t$ is the closest one to $\frac{1}{2}$.

$k$ is an integer close to $\frac{n}{9}$, i prefer to leave it to the others to determine which one.