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.

You almost did the problem yourself, just left out a few bits:
Assume, without loss of generality,
$$a_1 \geq a_2 \geq \dots \geq a_n$$
By the rearrangement inequality,
$$\implies a_1^2 + a_2^2 \dots a_n^2 \geq a_1a_2 + a_2a_3 \dots a_{n-1}a_n + a_na_1$$
Clearly, the required will be minimum if equality holds true above. 
This will happen when 
$$a_1 = a_2 = a_3 = \dots  = a_n = a$$
This means, the maximum value of the inequality is:
$$\sum_{i = 1}^n a^2 - 2a^2 = -na^2$$
Since, from the first equation, 
$$na = 1$$
$$a = \frac{1}{n}$$
Therefore, the minimum is, 
$$-n\cdot\left(\frac{1}{n^2}\right) = -\frac{1}{n}$$
