If the sum of $n$ cubes is zero, then the sum must be no larger than $\frac n3$. Assume that $a_1,...a_n$ are real numbers and $-1 \leq a_i\leq 1$ for $1\leq i\leq  n$. If $$a_1^3+\ldots +a_n^3=0$$ Then show that $$a_1+a_2+\ldots+a_n\le n/3$$ 
I just came cross this problem the other day. I totally have no idea how to prove it. Are there any hints about which theorem, principle or method to apply?
 A: This problem seems to be wrongly stated.  Suppose that $a_1,\ldots,a_{27} = 1/3$ and $a_{28} = -1$.  Then $\sum a_i^3 = 27/27 - 1 = 0$, but $\sum a_i = 27/3 - 1 = 8 \gt \frac{3}{28}$.
Perhaps the intended bound was $n/3$ rather than $3/n$?
A: Note $\let\leq\leqslant\let\geq\geqslant$that this is a question (day 2 problem 3) from the IMC (International Mathematics Competition for University Students) 1999.
The most elegant solution I know is the following: (It turns out that this is just the official solution, so (down)vote accordingly.)
For $x\geq-1$, $$x^3-\frac34x+\frac14=(x+1)\left(x-\frac12\right)^2\geq0.$$ Hence $$(a_1^3+\cdots+a_n^3)-\frac34(a_1+\cdots+a_n)+\frac n4\geq0,$$ that is, $$a_1+\cdots+a_n\leq\frac n3.$$

You could have come up with this solution if only you had the idea of considering an appropriate cubic $x^3+ax+b$ (with the above strategy in mind) which satisfies


*

*It is positive for $-1\leq x\leq 1$.

*If we sum up for $a_1,\ldots,a_n$ we should obtain $a_1+\cdots+a_n\leq\frac n4$. This means it is necessary that $a=-3b$.

*You could guess another condition such as that it has a factor $x+1$ (perhaps because a condition $a\geq-1$ in an equality is unusual). This gives $b=\frac14$.
Hence a possible cubic would be $x^3-\frac34x+\frac14$. Indeed, it works.
Note: examining the equality case should also help you find the appropriate cubic. Observe that equality holds for $n=9k$ and $8k$ of the numbers are $\frac12$, $k$ of them are $-1$. This tells us what the roots of the cubic should be.
