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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

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  • $\begingroup$ Tips: 1) First try the problem with $2014$ replaced with $3.$ 2) Interpret the condition $x_1+x_2+x_3=0$ as the $x_i$ lying in a plane, and you want to maximize $x_1^2+x_2^2+x_3^3$ -- the square of the distance between the plane and the origin. Use geometric ideas! $\endgroup$ – Ragib Zaman Aug 1 '14 at 9:44
  • $\begingroup$ I have konw if $a+b+c=0$,then $a^3+b^3+c^3=3abc$ $\endgroup$ – china math Aug 1 '14 at 9:54
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We want to maximize a continuous function on a compact set, so the maximum surely is attained. Let $(x_1,\ldots,x_{2014})$ be a point that maximizes the target function. If there are two indices $i\ne j$ with $x_i,x_j\notin\{-11,5\}$, then $x_i^2+x_j^2$ can be increased to $(x_i+h)^2+(x_j-h)^2=x_i^2+x_j^2+2h^2+2h(x_i-x_j)>x_i^2+x_j^2$ (if $|h|$ is small enough and $h$ has the same sign as $x_i-x_j$ - or arbitrary sign if $x_i=x_j$). Therefore, we can conclude that at most one coordinate is not on the boundary. So we have $m$ times $-11$, i.e. wlog $x_1=\ldots=x_m=-11$, then $2013-m$ times $5$, i.e. wlog. $x_{m+1}=\ldots=x_{2013}=5$, and one remaining value $x_{2014}=m\cdot11-(2013-m)\cdot 5=16m-10065$. For this to be $\in[-11,5]$, we need $m=629$, which makes $x_{2014}=-1$. So the target value is $$629\cdot(- 11)^2+1384\cdot 5^2 +(-1)^2. $$

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Another way - as the objective function is convex, at maximum we must have $n-1$ of the $x_i \in \{-11, 5\}$. So let $k$ of the variables be $-11$ and the remaining $2013-k$ be $5$ at maximum. We then have the constraint $$-11k + 5(2013-k)+x_{2014}=0 \implies x_{2014} = 16k-10065$$

So the objective function is $121k+25(2013-k)+(16k-10065)^2$, which gets maximised when $k=629$ and has a maximum value $110710$.

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  • $\begingroup$ There are (2013-k) 5 (ideally, there should be (2014-k) 5, but this condition won't lead to an integer solution), so your equation should be corrected and so should the maximum. $\endgroup$ – Huang Aug 1 '14 at 10:14
  • $\begingroup$ @Huang not sure what equation you think should be corrected. Are you also saying $110710$ is not the maximum? $\endgroup$ – Macavity Aug 1 '14 at 10:17
  • $\begingroup$ Sorry,. You have edited it. Originally it's -11k+ 5(2014-k) + x_2014 =0. I'm reading the post on my phone, and I can't see the updates immediately. $\endgroup$ – Huang Aug 1 '14 at 10:27

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