WakeUp wrote: 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 \[a_1^2+a_2^2+\ldots+a_{59}^2\le 2006\] Put $x_i=a_i+2$. Note that $x_i\in [0,19]$ and $\sum x_i=118$. We must to prove that $\sum x_i^2\le 4\cdot 59+2006=19\cdot 118=19\sum x_i$. But this inequality clearly holds because $x_i^2\leq 19\cdot x_i$ for all $1\leq i\leq 59$.

WakeUp wrote: 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 \[a_1^2+a_2^2+\ldots+a_{59}^2\le 2006\] $2006$ is not the best bound. We can prove: $a_1^2+...+a_{59}^2\le 1946<2006$

Proof

WakeUp wrote: 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 \[a_1^2+a_2^2+\ldots+a_{59}^2\le 2006\] we have : $x^{2}\leq\ 15x+34 , \forall\ x\in\ [-2,17]$ so: $\sum_{i=1}^{59}a_{i}^2\leq\ 15(\sum_{i=1}^{59}a_{i})+\sum_{i=1}^{59}34=2006$ anyway..karamata's inequality still stronger and gives the best bound!

Simple inequality (158)： Let $a_1,a_2,\cdots,a_{80}$ be positive integers such that $a_1+a_2+\cdots+a_{80}=123$ , prove that\[a^2_1+a^2_2+\cdots+a^2_{80}\le 2015.\]China Northern Mathematical Olympiad 2013： If $a_1,a_2,\cdots,a_{2013}\in[-2,2]$ and $a_1+a_2+\cdots+a_{2013}=0$ , find the maximum of $a^3_1+a^3_2+\cdots+a^3_{2013}$.

China: $(x-2)(x+1)^2\le 0\implies \sum a^3\le 3\sum a+2\cdot 2013=2\cdot 2013$ with equality iff $2013/3$ numbers are equal to $2$ and rest of them are equal to $-1$

MathUniverse wrote: $f(x)=x^2$ is convex function so using karamata's inequality, we get: $f(a_1)+...+f(a_{59})\le 6\cdot f(17)+f(2)+52\cdot f(-2)=1946$ I need help. What sequences should be considered, to get this inequality?

WakeUp wrote: 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 \[a_1^2+a_2^2+\ldots+a_{59}^2\le 2006\] $\color{yellow}{\clubsuit} \color{red}{\textit{\textbf{Proof:}}}$ We have $(x_i+2)(x_i-17)\le 0, \implies x_i^2\le 15x_i+ 34$, thus \[\sum_{i=1}^{59} x_i^2\le \sum_{i=1}^{59} (15x_i+ 34)=2006. \quad \blacksquare\]