How find this minimum $\sum_{i=1}^{n}a^2_{i}-2\sum_{i=1}^{n}a_{i}a_{i+1},a_{n+1}=a_{1}$ let $a_{1},a_{2},\cdots,a_{n}\ge 0$,and such $a_{1}+a_{2}+\cdots+a_{n}=1$.
Find this follow minimum 
$$I=a^2_{1}+a^2_{2}+\cdots+a^2_{n}-2a_{1}a_{2}-2a_{2}a_{3}-\cdots-2a_{n-1}a_{n}-2a_{n}a_{1}$$
My try:since
$$a^2_{1}+a^2_{2}+\cdots+a^2_{n}\ge a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n-1}a_{n}+a_{n}a_{1}$$
this is true because
$$\Longleftrightarrow \dfrac{1}{2}[(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+\cdots+(a_{n-1}-a_{n})^2+(a_{n}-a_{1})^2]\ge0$$
so
\begin{align*}
&a^2_{1}+a^2_{2}+\cdots+a^2_{n}-2a_{1}a_{2}-2a_{2}a_{3}-\cdots-2a_{n-1}a_{n}-2a_{n}a_{1}\\
&\ge a^2_{1}+a^2_{2}+\cdots+a^2_{n}-2(a^2_{1}+a^2_{2}+\cdots+a^2_{n})\\
&=-(a^2_{1}+a^2_{2}+\cdots+a^2_{n})
\end{align*}
if we use Cauchy-Schwarz inequality,we have
$$(a^2_{1}+a^2_{2}+\cdots+a^2_{n})(1+1+\cdots+1)\ge (a_{1}+a_{2}+\cdots+a_{n})^2=1$$
But this is not usefull.
and Then I can't, yesterday I have ask this problem：How find this inequality minimum $\sum_{i=1}^{n}a^2_{i}-2\sum_{i=1}^{n-1}a_{i}a_{i+1}$,
and the Ewan Delanoy use nice method to solve it.I don't know this problem have this nice methods to solve it too(maybe can use previous question methods?but I can't use it.) and this problem have someone have research? if no,I think this is nice problem.
and Now I guess This problem when $a_{i}=\dfrac{1}{n}$ then $I$ is minimum?and the minimum is
$$I_{min}=-\dfrac{1}{n}?$$
Thank you 
 A: UPDATE 11/14/2013 : Part of the first proposed answer was wrong. The method
proposed only works for $n<7$, and I currently have no complete solution for $n\geq 7$.  All that is corrected in the updated version below.
Let us put
$$
Q_n(a_1,a_2,\ldots ,a_n)=
a^2_{1}+a^2_{2}+\cdots+a^2_{n}-2a_{1}a_{2}-2a_{2}a_{3}-\cdots-2a_{n-2}a_{n-1}-2a_{n-1}a_{n}-2a_na_1
$$
and
$$
T_n(a_1,a_2,a_3,\ldots,a_n)=
Q_n(a_1,a_2,\ldots,a_n)+\frac{(a_1+a_2+a_3+\ldots +a_n)^2}{n}
$$
For $3\leq n\leq 6$, the minimum is $-\frac{1}{n}$ (attained when all coordinates
are equal to $\frac{1}{n}$), because of
$$
\begin{array}{lcl}
T_6(a_1,a_2,\ldots,a_6)&=&
\frac{1}{42}\bigg(-5a_1+a_2+a_3+a_4-5a_5+7a_6\bigg)^2 \\
& & +\frac{1}{28}\bigg(-3a_1+2a_2+2a_3-5a_4+4a_5\bigg)^2
+\frac{1}{4}\bigg(-a_1+2a_2-2a_3+a_2\bigg)^2 \\
T_5(a_1,a_2,\ldots,a_5) &=& 
\frac{1}{30}\bigg(-4a_1+a_2+a_3-4a_4+6a_5\bigg)^2
+\frac{1}{6}\bigg(-a_1+a_2-2a_3+2a_4\bigg)^2 \\
& &
+\frac{1}{2}\bigg(-a_2+a_3\bigg)^2+\frac{1}{2}\bigg(-a_1+a_2\bigg)^2 \\
& & \\
T_4(a_1,a_2,a_3,a_4) &=& 
\frac{1}{20}\bigg(-3a_1+a_2-3a_3+5a_4\bigg)^2
+\frac{1}{20}\bigg(-a_1-3a_2+4a_3\bigg)^2 \\
& &
+\frac{3}{4}\bigg(-a_1+a_2\bigg)^2 \\
& & \\
T_3(a_1,a_2,a_3) &=& 
\frac{1}{3}\bigg(-a_1+a_2+2a_3\bigg)^2
+\bigg(-a_1+a_2\bigg)^2 \\
\end{array}
$$
Unfortunately, this method does not work any more for $n \geq 7$. Indeed, in that
case the minimum is $\leq -\frac{1}{6}$ (because 
$Q_n(0,\ldots,0,\frac{1}{6},\frac{1}{3},\frac{1}{3},\frac{1}{6})=
-\frac{1}{6}$) ; on the other hand, the polynomial
$$
R_n(a_1,a_2,\ldots ,a_n)=Q_n(a_1,a_2,\ldots,a_n)+\frac{(a_1+a_2+a_3+\ldots +a_n)^2}{6}
$$
is not nonnegative on ${\mathbb R}^n$ any more. For example, for $n=7$ we have
$$
R_7(4, 1, 0, -1, 0, 3, 5)=(-2) < 0
$$
