Prove for non-zero $a_k$ satisfying $\sum a_k=0$ there exists a permutation such that $a_1a_2+a_2a_3+a_3a_4+\cdots+a_{n-1}a_{n}+a_{n}a_1 \lt 0$. Let $b_1,b_2,\dots,b_n$ denote non-zero real numbers satisfying $\sum_{i=1}^n {b_i}=0$. Prove that there exists a permutation $ a_1,a_2,\dots,a_n$ of these numbers such that $a_1a_2+a_2a_3+a_3a_4+\cdots+a_{n-1}a_{n}+a_{n}a_1 \lt 0$.
I'm very grateful for your guidance in solving this problem.
 A: What do you get when you sum over all possible permutations?
You get $$2n(n-2)!\sum\limits_{1\leq i<j\leq n}a_ia_j=2n(n-2)!\dfrac{(a_1+a_2+\dots+a_n)^2-\sum\limits_{i=1}^na_i^2}{2}=-n(n-2)!\sum\limits_{i=1}^na_i^2.$$
Since this sum is negative at least one sumand is negative.
A: We consider a random permutation of $b_1,b_2,b_3,\cdots,b_n$ 
 like $a_1,a_2,\cdots,a_{n} $ and define $X_i=a_ia_{i+1},X_{n}=a_{n}a_1$.
$$E[X_i]=\sum_{(v,w)}{b_vb_w}\times P(a_{i}={b_v},a_{i+1}={b_w})$$
$$=\sum_{(v,w)}{b_vb_w\times{\frac{(n-2)!}{n!}}} \ \ \ \ \ \ \ \ \ \ $$
Now consider $X=\sum_{i=1}^{n}{X_{i}}$ then 
$$E[X]=\sum_{i=1}^{n}{E[X_i]}=n\sum_{(v,w)}{b_vb_w\times \frac{1}{n(n-1)}} $$$$=\frac{1}{n-1}\sum_{(v,w)}{b_vb_w}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$
As regards $\sum_{i=1}^{n}{b_i}=0$ then $(\sum_{i=1}^{n}{b_i})^{2}=\sum_{i=1}^{n}({b_i})^{2}+\sum_{(v,w)}{b_vb_w}=0 $ 
so $\ \sum_{(v,w)}{b_vb_w}\lt {0} \ $ so $E[X]\lt {0} $ then we should find a permutation that 
$a_1a_2+a_2a_3+a_3a_4+\cdots+a_{n-1}a_{n}+a_{n}a_1 \lt 0$.
