Let $x_1, x_2, \dots, x_n$ be sequence of integers such that $\dots$

Question

Question:
Let $x_1, x_2, \dots, x_n$ be sequence of integers such that

$-1 \leq x_i \leq 2$ for $i = 1, 2, \dots, n$.
$x_1 + x_2 + \dots + x_n = 19$
${x_1}^2 + {x_2}^2 + \dots + {x_n}^2 = 99$

Determine the minimum and maximum possible values of $${x_1}^3 + {x_2}^3 + \dots + {x_n}^3$$

Attempt:
To simply the problem, the domain $-1 \leq x_i \leq 2$ is simplified to integers $-1, 1, 2$ since $0$ is of no use other than elongate $n$ to $\infty$. Also pairs of both positive would be impossible since (2) above would overshoot $19$ to meet the requirements of (3). Thus we are left with the following options, $(-1, 1)$ and/or $(-1, 2)$. To determine the quantity/plausibility of each of our options, we determine the number of $-1$ and $1$ for $(-1, 1)$ and so on.

For (-1, 1), Let $x$ be the number of $-1$ and $y$ be the number of $1$: $$x+y = 99; -x + y = 19$$ $$x = 40, y = 59$$

For (-1, 2), we have no integer solution, thus we are left with the $(-1, 1)$.

That being said, ${x_1}^3 + {x_2}^3 + \dots + {x_n}^3$, would be $(-1)^3*40 + 1*59 = 19$ for both maximum and minimum values.

Can someone check if my reasoning is correct.

@vadim123 possible values of $x_i$ – Joey Andres Aug 09 '14 at 15:00 — Joey Andres, Aug 09 '14 at 15:00
Why can't the sequence contain all three values? – aschepler Aug 09 '14 at 15:01 — aschepler, Aug 09 '14 at 15:01
@aschepler crap, forgot that case. – Joey Andres Aug 09 '14 at 15:03 — Joey Andres, Aug 09 '14 at 15:03
I think it makes more sense now, cheers. – Joey Andres Aug 09 '14 at 15:04 — Joey Andres, Aug 09 '14 at 15:04

score 1 · Answer 1 · answered Aug 09 '14 at 16:20

It is best to work with a rescaled version of the problem. Put $y_i=x_i+1$.

Then all the $y_i$s belong to the $[0,3]$ interval and: $$\sum y_i = 19+n,\qquad \sum y_i^2 = 99+2\sum x_i+n = 137+n.$$

Due to the AM-QM inequality, we must have $n\geq 4$. Under such assumption, the convexity of $\sum x_i^3$ implies that the maximum is on the boundary (that is the intersection between a sphere and a cube), while the minimum can be found through Lagrange multipliers, for instance. The inequality between the cubic mean and the quadratic mean gives: $$\sum y_i^3\geq n\left(1+\frac{137}{n}\right)^{3/2},$$ from which it follows that: $$\sum x_i^3\geq n\left(1+\frac{137}{n}\right)^{3/2}-3(137+n)-3(19+n)-n.$$ However, the last inequality is non-trivial only for $n\leq 10$.

Let $x_1, x_2, \dots, x_n$ be sequence of integers such that $\dots$

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