Can the Inequality of the Sum be Proven Using the Cube Root of -1?

[anemone]

Assume that $x_1,\,x_2,\,\cdots,\,x_n\ge -1$ and $\displaystyle \sum_{i=1}^n x_i^3=0$. Prove that $\displaystyle \sum_{i=1}^n x_i\le \dfrac{n}{3}$.

 

[Euge]

Interesting problem. Here is my solution.

Spoiler

It suffices to show that $\max\limits_{x\in [-1,\infty)} (x - \frac43 x^3) = \frac13$, since then we may estimate $$\sum x_i = \sum x_i - \frac{4}{3}\sum x_i^3 = \sum \left(x_i - \frac{4}{3}x_i^3\right) \le \sum \frac{1}{3} = \frac{n}{3}$$ Note that for all $x \ge -1$, \[\begin{align}\frac13 - \left(x - \frac43 x^3\right) &= \frac{4x^3 - 3x + 1}{3}\\ &= \frac{4[(1 + x)^3 - 3(1 + x)^2 + 3(1 + x) - 1] - 3(1 + x) + 4}{3}\\ &= \frac{4(1 + x)^3 - 12(1 + x)^2 + 9(1 + x)}{3}\\ &= \frac{(1 + x)[2(1 + x) - 3]^2}{3}\end{align}\] is nonnegative, and equals zero if $x = -1$ or $x = \frac12$, proving the claim.

 

[anemone]

Thanks for your participation, in which your solution is almost the same as the official solution, very well done, Euge!

 

[DaalChawal]

Interesting problem. Here is my solution.

Spoiler

It suffices to show that $\max\limits_{x\in [-1,\infty)} (x - \frac43 x^3) = \frac13$, since then we may estimate $$\sum x_i = \sum x_i - \frac{4}{3}\sum x_i^3 = \sum \left(x_i - \frac{4}{3}x_i^3\right) \le \sum \frac{1}{3} = \frac{n}{3}$$ Note that for all $x \ge -1$, \[\begin{align}\frac13 - \left(x - \frac43 x^3\right) &= \frac{4x^3 - 3x + 1}{3}\\ &= \frac{4[(1 + x)^3 - 3(1 + x)^2 + 3(1 + x) - 1] - 3(1 + x) + 4}{3}\\ &= \frac{4(1 + x)^3 - 12(1 + x)^2 + 9(1 + x)}{3}\\ &= \frac{(1 + x)[2(1 + x) - 3]^2}{3}\end{align}\] is nonnegative, and equals zero if $x = -1$ or $x = \frac12$, proving the claim.

Is there any reason why we took 4/3 to be multiplied by x^3? I know it simplifies nicely after that. Is the reason because -1 will then be its root?

 

[Euge]

Is there any reason why we took 4/3 to be multiplied by x^3? I know it simplifies nicely after that. Is the reason because -1 will then be its root?

Yes indeed, there is a reason:

Spoiler

My idea was to find $\alpha$ such that $x - \alpha x^3$ has maximum $\frac13$ at some point in $[-1,\infty)$, and $\alpha = \frac43$ accomplishes that. We would then have $x_i - \frac{4}{3}x_i^3 \le \frac{1}{3}$ for each $i$ so that $\sum x_i = \sum (x_i - \frac{4}{3}x^3)$ must be no greater than $\frac{n}{3}$.