# Inequality involving an absolute value

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 $$k$$, $$k, such that: $$\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.

• 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 $k$! Commented Apr 8 at 8:20

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.

• 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? Commented Apr 7 at 18:46
• 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? Commented Apr 7 at 20:09
• @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.
– chi
Commented Apr 7 at 20:59
• @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 $(x + 4)(x - b)^2 \ge 0$. Commented Apr 8 at 0:51
• @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 $k$. Commented Apr 8 at 1:06