It was posted here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=548178

Thank arqady .

I've solved it when $x_i\in[-1,1]$. Hongbing Yu have given the problem to me.

Pham Kim Hung - Secrets in Inequalities (volume 1)： If $a_1,a_2,\cdots,a_{n}\in[-1,1]$ and $a_1+a_2+\cdots+a_{n}=0$ , find the maximum of $a^3_1+a^3_2+\cdots+a^3_{n}$. Baltic Way 2006: Suppose that the real numbers $a_i\in [-2,17],\ i=1,2,\ldots,59,$ satisfy $a_1+a_2+\ldots+a_{59}=0.$ Prove that $a_1^2+a_2^2+\ldots+a_{59}^2\le 2006.$

sqing wrote: If $a_1,a_2,\cdots,a_{2013}\in[-2,2]$ and $a_1+a_2+\cdots+a_{2013}=0$ , find the maximum of $a^3_1+a^3_2+\cdots+a^3_{2013}$. Observe that $ x^3 \geq -2+3x $ is equivalent to $ (x+1)^2(x+2) \geq 0 $ which is true for $x \geq -2$ and $ 2+3x \geq x^3 $ is equivalent to $ (x+1)^2(2-x) \geq 0$ which is true for $ 2 \geq x$ Since, $a_1,a_2,\cdots,a_{2013}\in[-2,2]\quad $, we have $$ 2+3a_1\quad \geq \quad a_1^3 \quad \geq \quad -2+3a_1$$ $$2+3a_2\quad \geq \quad a_2^3\quad \geq\quad -2 + 3a_2$$ $$2+3a_3 \quad \geq \quad a_3^3 \quad \geq \quad -2+3a_3$$ $$\cdots $$ $$2+3a_{2013}\quad \geq \quad a_{2013}^3\quad \geq \quad -2 + 3a_{2013}$$ Adding the all the above inequalities, we get $$2\cdot2013 +3\left(a_1+a_2+ \cdots +a_{2013}\right)\geq a_1^3 + a_2^3 + a_3^3 + \cdots + a_{2013}^3 \geq -2\cdot 2013 +3\left(a_1 + a_2 + \cdots + a_{2013}\right)$$ $$\Leftrightarrow 4026 \geq a_1^3 + a_2^3 + a_3^3 + \cdots + a_{2013}^3 \geq -4026$$ Therefore, the maximum value of $a_1^3 + a_2^3 + a_3^3 + \cdots + a_{2013}^3$ is $4026$ and minimum value of $a_1^3 + a_2^3 + a_3^3 + \cdots + a_{2013}^3$ is $-4026$

Let $a_1,a_2,\cdots,a_{80}$ be positive integers such that $a_1+a_2+\cdots+a_{80}=123$ . Prove that\[a^2_1+a^2_2+\cdots+a^2_{80}\le 2015.\]Let $-1\le x_i \le1$ and there are 31 numbers such that $x_1+x_2+...+x_{31}=28$. Prove that $${x^2_1}+{x^2_2}+\ldots+{x^2_{31}}\le30$$