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Given the numbers $x_1, x_2,...,x_n$ in the interval $[-4;4]$, such that $ \sum_{i=1}^{n} x_i=0$, prove that: $$|\sum_{i=1}^{n} x_i^3|\le 16n$$ I have tried using various known inequalities, namely the Minkowsky- and the AM-GM inequalties, but I ran into a false statement. What I know for sure is that there is an integer $k$, $k<n$, such that: $$\sum_{i=1}^{k} x_i=-\sum_{i=k+1}^{n} x_i$$ I also tried to express the summation using the fact that the sum of the cubes equals the sum cubed minus the products, but since the sum cubed equals zero, then $\sum_{i=1}^{n} x_i^3=$-(the products resulting from cubing the sum), but I could not find a formula for the products, which would help me in any way.

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  • $\begingroup$ Re: "there is an integer $k$, $k<n$, such that: $\sum_{i=1}^{k} x_i=-\sum_{i=k+1}^{n} x_i$": Indeed, that statement is true for every $k$! $\endgroup$
    – ruakh
    Commented Apr 8 at 8:20

1 Answer 1

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We have $(4 - x)(x + 2)^2 \ge 0$ for all $x \in [-4, 4]$ which results in $x^3 \le 12x + 16$ for all $x\in [-4, 4]$. Thus, we have $$\sum_{i=1}^n x_i^3 \le \sum_{i=1}^n 12x_i + 16n = 16n.$$

We have $(x + 4)(x - 2)^2 \ge 0$ for all $x \in [-4, 4]$ which results in $x^3 \ge 12x - 16$ for all $x\in [-4, 4]$. Thus, we have $$\sum_{i=1}^n x_i^3 \ge \sum_{i=1}^n 12x_i - 16n = -16n.$$

Thus, we have $-16n \le \sum_{i=1}^n x_i^3 \le 16n$.

We are done.

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    $\begingroup$ Thank you for the solution! Analysing the solution I have got a question related to similar problems. Are we dead-lost without that observation involving $(4-x)(2+x)$ etc? Is there a way to make such observations in a consequent way? Or it happens only by experience? $\endgroup$
    – fikooo
    Commented Apr 7 at 18:46
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    $\begingroup$ I second fikooo's follow-up question. I was just thinking: how did you come up with this answer? Probably just lots of experience with polynomials? $\endgroup$ Commented Apr 7 at 20:09
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    $\begingroup$ @fikooo Mine is only a guess, but I believe that the wanted inequality and the hypotheses somehow suggest to search for inequalities of the form $x^3 \leq k\cdot x + 16$ for some $k$, so that summing over all the $x$'s we can conclude. It's not that obvious at first, but I guess this might have been the initial guess that lead to this answer. $\endgroup$
    – chi
    Commented Apr 7 at 20:59
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    $\begingroup$ @fikooo It is the experience. As chi pointed out, we want the form of $x^3 \le kx + 16$ etc. We hope $x^3 \le kx + 16$ is the expanding of something non-negative. Since it is cubic, it should be something like $(x+a)(x -b)^2 \ge 0$ for $x\in [-4, 4]$. Then we consider $(4-x)(x-b)^2 \ge 0$ or $(x + 4)(x - b)^2 \ge 0$. $\endgroup$
    – River Li
    Commented Apr 8 at 0:51
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    $\begingroup$ @fikooo Actually constructing $(x - a)(x - b)^2 \ge 0$ is a trick which are used for some problems by users in MSE. For example, in my answer, I used the trick as follows: We have $$\left(a_k - \frac{n - 2}{n}\right)^2(a_1 - a_k) \ge 0$$ which results in $$a_k^3 \le \left(2 + a_1 - \frac{4}{n}\right)a_k^2 + \left(\frac{4}{n} - 1 + \frac{4a_1}{n} - \frac{4}{n^2} - 2a_1\right)a_k + \frac{(n - 2)^2a_1}{n^2}.$$ Then take a sum over $k$. $\endgroup$
    – River Li
    Commented Apr 8 at 1:06

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