Question:
let $x_{i}\in[-11,5],i=1,2,\cdots,2014$,and such $$x_{1}+x_{2}+\cdots+x_{2014}=0$$ find the maximum of the value $$x^2_{1}+x^2_{2}+\cdots+x^2_{2014}$$
since $$(x_{i}-5)(x_{i}+11)\le 0$$ then $$x^2_{i}\le -6x_{i}+55$$ so $$x^2_{1}+x^2_{2}+\cdots+x^2_{2014}\le -6(x_{1}+\cdots+x_{n})+55\cdots 2014=2014\cdot 55$$
But I find this is not maximum the value,because Now I found this inequality can't equalient,
idea 2: let $$x_{i}+3=a_{i}\in[-8,8]$$ then $$a_{1}+a_{2}+\cdots+a_{2014}=x_{1}+x_{2}+\cdots+x_{2014}+3\times 2014=6042$$ and then we $$\sum_{i=1}^{2014}x^2_{i}=\sum_{i=1}^{2014}(a_{i}-3)^2=\sum_{i=1}^{2014}a^2_{i}-6\sum_{i=1}^{2014}a_{i}+9\cdot 2014=\sum_{i=1}^{2014}a^2_{i}-18126$$ so How find it? Thank you