find the range $(x_{1}+x_{2}+\cdots+x_{2009})^2=4(x_{1}x_{2}+x_{2}x_{3}+\cdots+x_{2009}x_{1})^2$ let $x_{i}\ge 0(i=1,2,3,\cdots,2009$,and such $$(x_{1}+x_{2}+\cdots+x_{2009})^2=4(x_{1}x_{2}+x_{2}x_{3}+\cdots+x_{2009}x_{1})=4$$
find the range  $\sum_{i=1}^{2009}x^2_{i}$
I guess the range is $[1.5,2]$ 
because I think when $x_{1}=1,x_{2}=1,x_{3}=x_{4}=\cdots=x_{2009}=0$then is maximum of value $2$
and when $x_{1}=0.5,x_{2}=1,x_{3}=0.5,x_{4}=\cdots=x_{2009}=0$ is minmum of the value  $1.5$
 A: I think you are right!
We'll prove that
$$(x_1+x_2+...+x_{2009})^2\geq4(x_1x_2+x_2x_3+...+x_{2009}x_1)$$ for all non-negatives $x_i$.
Indeed, let $x_2=\max\limits_{i}\{x_i\}$.
Thus, $$x_2x_{2009}\geq x_1x_{2009}$$ and by AM-GM we obtain:
$$4(x_1x_2+x_2x_3+...+x_{2009}x_1)\leq4(x_1+x_3+...+x_{2009})(x_2+x_4+...+x_{2008})\leq\left(\sum\limits_{i=1}^{2009}x_i\right)^2,$$
where the equality occurs for
$$x_1(x_4+x_6+...+x_{2008})=0,$$ 
$$x_2(x_5+...+x_{2007})=0,$$ which gives
$$x_5=...=x_{2007}=0.$$ 
Now, if $x_1=0$ then $x_2x_3=1,$ which gives 
$$2=\sum_{i=1}^{2009}x_i\geq x_2+x_{3}\geq2\sqrt{x_2x_{3}}=2,$$
which gives $x_2=x_{3}=1$, $x_1=x_4=...=x_{2009}=0$ and  $\sum\limits_{i=1}^{2009}x_i^2=2$.
If $x_1>0$ we obtain:
$$x_1(x_4+...+x_{2008})=0,$$ 
which gives $x_4=x_5=...=x_{2008}=0.$
Now, let $x_2=a$, $x_1=b$,$x_3=c$ and $x_{2009}=d$.
Thus, $$(a+b+c+d)^2=4(ab+bd+ca)=4,$$
which gives $a+b+c+d=2$ and $ab+bd+ca=1.$
Thus,  by AM-GM  $$1=ab+bd+ca=ab+bd+dc+ca-dc=(a+d)(b+c)-dc\leq$$
$$\leq\left(\frac{a+b+c+d}{2}\right)^2-dc=1-dc\leq1,$$
which gives $dc=0$ and $a+d=b+c=1.$
Now, let $c=0$. 
Thus, $b=1=a+d$ and since $a\geq b$, we obtain $d=0$ and $\sum\limits_{i=1}^{2009}x_i^2=2$ again.
Let $d=0$.
Id est, by C-S 
$$\sum_{i=1}^{2009}x_i^2=a^2+b^2+c^2=1+\frac{1}{2}(1+1)(b^2+c^2)\geq1+\frac{1}{2}(b+c)^2=1.5.$$
The equality occurs for $a=1$ and $b=c=\frac{1}{2},$ which says that we got a minimal value.
Also, $$\sum_{i=1}^{2009}x_i^2=a^2+b^2+c^2=1+(b+c)^2-2bc\leq1+(b+c)^2=2.$$
The equality occurs for $a=b=1$ and $c=0,$ which says that we got a maximal value.
A: Solution
We are to state a lemma, on which the whole thing is based.

Lemma
Let $n > 4$ and $x_0,x_1,\ldots,x_n,x_{n+1}$ be non-negative real
  numbers, with $x_0=x_n$ and $x_{n+1} = x_1$. Then $$\left (\sum_{i=1}^n x_i \right )^2 \geq 4\sum_{i=1}^n
x_{i}x_{i+1},$$
  with the equality holding if and only if, for any $j~~ (1\leq j\leq n)$ we have$$x_j =\dfrac {1}{2} \sum\limits_{i=1}^n x_i = x_{j-1}+x_{j+1}$$ and the rest
  of the variables are equal to $0$.

Come back to the present case. Since $n=2009> 4$,  $\sum\limits_{i=1}^n x_i = 2$ and $\sum\limits_{i=1}^n x_{i}x_{i+1} = 1$, thus the equality holds. Therefore, for any $j~~~(1 \leq j \leq 2009)$, we have $x_j=1$,$x_{j-1}+x_{j+1}=1,$ and the rest are equal to $0$.
Under these constraints, we obtain $$S = \sum_{i=1}^n x_i^2 = x_j^2 + x_{j-1}^2 + x_{j+1}^2 =1 + x_{j-1}^2 + x_{j+1}^2,$$ where $x_{j-1}, x_{j+1} \geq 0$ and $x_{j-1}+x_{j+1}= 1$. Thus, $\dfrac {1}{2} \leq x_{j-1}^2 + x_{j+1}^2 \leq 1$, therefore $$\dfrac {3}{2} \leq S \leq 2.$$ 
