Maximizing $x_1+x_2+\cdots+x_{10}$, where each $x_i$ lies between $-1$ and $1$, and $x_1^3+x_2^3+\cdots+x_{10}^3=0$

Question

Let $x_1, x_2, \ldots, x_{10}$ be ten quantities each lying between $-1$ and $1$, and the sum of cubes of these ten quantities is zero. Find the maximum value of $x_1+x_2+\cdots+x_{10}$.

I have tried substituting $x_1=\sin(y_1)$, $x_2=\sin(y_2)$, and so on, and then used the identity for $\sin 3x$ to simplify the cubes leading to
$$x_1 + x_2 + \cdots x_{10} = \frac13 (\sin 3y_1 + \sin 3y_2 +\cdots + \sin 3y_{10})$$ Now is the statement $$\frac13 (\sin 3y_1 + \sin 3y_2 +\cdots + \sin 3y_{10}) \leq \frac{10}{3}$$ right? Is the answer $10/3$? How do I maximise $$\frac13 (\sin 3y_1 + \sin 3y_2 +\cdots + \sin 3y_{10})$$ under the constraints?

Answer 1

The answer is indeed a little lower than the upper bound of $10/3$. You can come close to having it all, but not quite have everything.

It's actually $9^{2/3}-1\approx 3.3267$.

Riding the S curve

Without loss of generality, we may assume that the $x$ values are in ascending order defined by $x_1\le x_2\le x_3\le ... \le x_{10}$. In that case the terms $x_1$ through $x_m$ are all negative for some whole number $m\in\{1,2,3,4,5,6,7,8,9\}$, and all subsequent terms are nonnegative.

Suppose, then, that $x_k$ and $x_{k+1}$ are two consecutive negative terms. Below we see what happens if we decrease $x_k$ and increase $x_{k+1}$ so that the sum of their cubes is constant:

Because of the curvature of the function for negative arguments, incrementing or decrementing the cubes equally will impart unequal changes to the uncubed terms: $x_{k+1}$ is increased more than $x_k$ is decreased and therefore the sum is increased by driving these terms farther apart. To maximize the contribution to the sum from the negative terms, therefore, we should drive those terms apart, pushing some of them all the way to $-1$ and others to $0$.

Now let's look at what happens with the nonnegative terms, including those that were pushed from negative to zero in accordance with the above strategy:

The cubing function now curves the opposite way, so pushing nonnegative terms farther apart will decrease rather than increase the sum of those terms. We should therefore equalize all nonnegative terms to maximize the sum.

So, to optimize both parts of the S curve, we should drive all negative terms to $-1$ and render all the remaining terms equal. Then with $m$ negative terms the remaining $10-m$ terms are given by

$x_k=[m/(10-m)]^{1/3}, k>m$

and the sum is rendered as

$f(m)=-m+(10-m)[m/(10-m)]^{1/3}=m^{1/3}(10-m)^{2/3}-m$

We then try out the possibilities for $m$, given above as $m\in\{1,2,3,4,5,6,7,8,9\}$. We discover that $m=1$ gives the maximum result, with a value given in the Spoiler.

Missed it by that much

Why did we come close to the upper bound of $10/3$ and yet ultimately we fell short? It was noted above that the number of negative terms had to be a whole number, thus constraining the candidates we could try. But what if we ignored this constraint? If $m$ were a continuous variable we would maximize $f(m)$ above by setting its derivative to zero.

Sure enough, the optimal value of $m$ from this method is not a whole number. It is the fractional number $10/9$. And for $m=10/9$, true to form, we hit $f(m)=10/3$. We meet the trigonometric upper bound after all ... but only by cheating. In the real world where $m$ has to be a whole number, we can only come close by selecting the nearest available whole number, and so we settle for $m=1$ and a sum value slightly less than our ideal one.

Answer 2

@RishiShekher What you have done is established an upper bound, to the problem that has to be attained. What the error in your argument is that you never explain whether equality can occur w.r.t the constraint, your argument is analogous to this. We all know the following inequality is true $\left(x+\frac{1}{x}\right)\ge 2$ for all $x\in \mathbb{R}$, but the same inequality could posed as $x+\frac{1}{x}\ge 1$. The difference is the first one, equality can be attained whereas for the second it can't.

P.S: This is not an answer just a comment that points out the error, the comment box couldn't handle the word count