How find this inequality minimum $\sum_{i=1}^{n}a^2_{i}-2\sum_{i=1}^{n-1}a_{i}a_{i+1}$ 
Let $n$ be a given positive integer, and let $a_{1},a_{2},\cdots,a_{n}\ge 0$ such that $a_{1}+a_{2}+\cdots+a_{n}=1$.
Find this  minimum value
$$a^2_{1}+a^2_{2}+\cdots+a^2_{n}-2a_{1}a_{2}-2a_{2}a_{3}-2a_{3}a_{4}-\cdots-2a_{n-2}a_{n-1}-2a_{n-1}a_{n}.$$

I think maybe we can use this well known
$$a^2_{1}+a^2_{2}+\cdots+a^2_{n}\ge a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n-1}a_{n}+a_{n}a_{1}?$$
But this problem is only
$$-2a_{1}a_{2}-2a_{2}a_{3}-2a_{3}a_{4}-\cdots-2a_{n-2}a_{n-1}-2a_{n-1}a_{n}$$
so I can't it.Thank you for help.
By the way:
I don't know this problem have Someone research?if No,I think this is interesting problem.

I show below that the minimum is $-\frac{1}{6}$ for $n \geq 4$, and $-\frac{1}{7}$ when $n =  3$.
If we put
$$
Q_n(a_1,a_2,\ldots ,a_n)=
a^2_{1}+a^2_{2}+\cdots+a^2_{n}-2a_{1}a_{2}-2a_{2}a_{3}-2a_{3}a_{4}-\cdots-2a_{n-2}a_{n-1}-2a_{n-1}a_{n}
$$
and
$$
R_n(a_1,a_2,\ldots ,a_n)=Q_n(a_1,a_2,\ldots,a_n)+\frac{(a_1+a_2+a_3+\ldots +a_n)^2}{6}
$$
then  for $n\geq 4$, we have the identity
$$
R_n(a_1,a_2,a_3,\ldots,a_n)=
\frac{1}{42}\bigg(\sum_{k=1}^{n-2}a_k-5a_{n-1}+7a_n\bigg)^2
+\frac{1}{28}\bigg(\sum_{k=1}^{n-3}2a_k-5a_{n-2}+4a_{n-1}\bigg)^2
+\frac{1}{4}\bigg(\sum_{k=1}^{n-4}2a_k-2a_{n-3}+a_{n-2}\bigg)^2
$$
Since $Q_n(0,\ldots, 0,\frac{1}{6},\frac{1}{3},\frac{1}{3},\frac{1}{6})=-\frac{1}{6}$,
we see that the minimum is $-\frac{1}{6}$ for $n\geq 4$.
For $n=3$, the minimum is $-\frac{1}{7}$ attained at
$(\frac{2}{7},\frac{3}{7},\frac{2}{7})$ (thanks Macavity) because of
$$
Q_3(a_1,a_2,a_3)+\frac{(a_1+a_2+a_3)^2}{7}=
\frac{1}{56}\bigg(a_1-6a_2+8a_3\bigg)^2
+\frac{1}{8}\bigg(-3a_1+2a_2\bigg)^2
$$
