Consider $n$ is a positive integer number. Let the set $I_k$, $1\leq k \leq n$, is defined by $I_k=\{1,2,\cdots ,n\}-\{k\}$. Assume that the polynomial $h_k$ is defined by: $$ h_k=\prod_{\substack { 1\leq t \leq n}} (a_k+b_t)\, \prod_{\substack { i,j\in I_k \\ i<j}} (a_i-a_j). $$ My question: How to proof the following relation: $$ \sum_{k=1}^n\, (-1)^{k-1}\, h_k=(\prod_{\substack { 1\leq i,j \leq n \\ i<j}} (a_i-a_j))\, (a_1+b_1+a_2+b_2+\cdots + a_n+b_n)\tag{1} $$
Example: Let $n=4$ then the polynomial $h_k$, $1\leq k \leq 4$, are given by: $$ \begin{array}{} h_1= \left( a_{{1}}+b_{{1}} \right) \left( a_{{1}}+b_{{2}} \right) \left( a_{{1}}+b_{{3}} \right) \left( a_{{1}}+b_{{4}} \right) \left( a_{{2}}-a_{{3}} \right) \left( a_{{2}}-a_{{4}} \right) \left( a_{{3}}-a_{{4}} \right),\\ h_2=\left( a_{{2}}+b_{{1}} \right) \left( a_{{2}}+b_{{2}} \right) \left( a_{{2}}+b_{{3}} \right) \left( a_{{2}}+b_{{4}} \right) \left( a_{{1}}-a_{{3}} \right) \left( a_{{1}}-a_{{4}} \right) \left( a_{{3}}-a_{{4}} \right),\\ h_3=\left( a_{{3}}+b_{{1}} \right) \left( a_{{3}}+b_{{2}} \right) \left( a_{{3}}+b_{{3}} \right) \left( a_{{3}}+b_{{4}} \right) \left( a_{{1}}-a_{{2}} \right) \left( a_{{1}}-a_{{4}} \right) \left( a_{{2}}-a_{{4}} \right),\\ h_4=\left( a_{{4}}+b_{{1}} \right) \left( a_{{4}}+b_{{2}} \right) \left( a_{{4}}+b_{{3}} \right) \left( a_{{4}}+b_{{4}} \right) \left( a_{{1}}-a_{{2}} \right) \left( a_{{1}}-a_{{3}} \right) \left( a_{{2}}-a_{{3}} \right). \end{array} $$ Then we can check that the relation $(1)$ holds for $n=4$, as follows $$ \begin{array}{} \sum_{k=1}^4\, (-1)^{k-1}\, h_k=\\ \left( a_{{1}}-a_{{2}} \right) \left( a_{{1}}-a_{{3}} \right) \left( a_{{1}}-a_{{4}} \right) \left( a_{{2}}-a_{{3}} \right) \left( a_{{2}}-a_{{4}} \right) \left( a_{{3}}-a_{{4}} \right) \left( a_{{1}}+b_{{1}}+\cdots +a_{{4}}+b_{{4} } \right). \end{array} $$ My try: If $a_i=a_j$ for some $1\leq i , j \leq n$, $i\neq j$ then the $\operatorname{RHS}$ of $(1)$ is zero and also we can prove the $\operatorname{LHS}$ of $(1)$ is zero ($n-2$ of $h_k$'s are zero and the other two $h_k$ are equal). I dont know how to show that if $\sum_{i=1}^n(a_i+b_i)=0$ then $\operatorname{LHS}$ of $(1)$ is zero.
Code: The Maple code of this question is provided here.
Thanks for any suggestions.
Well, here's a start. I think you'll find that $h_i$ has degree $k$ in $a_i$ and $k-2$ in $a_j$ for $j \ne i$, and total degree $k(k-1)/2+1$ in $a_1,\ldots,a_k$. Thus the left side of (1) has degree $k$ in each $a_j$ and total degree $k(k-1)/2+1$ in $a_1,\ldots,a_k$. So does the right side. The left side is $0$ when $a_i = a_j$ for some $i \ne j$, so it must be $\displaystyle\prod_{\substack { 1\leq i,j \leq n \\ i<j}} (a_i-a_j)$ times a polynomial that has degree $1$ in each $a_i$ and total degree $1$ in $a_1, \ldots, a_k$. Moreover, that polynomial is symmetric under permutations of the $a_i$, so it must be of the form $A \sum_i a_i + B$, where $A$ and $B$ are polynomials in $b_1,\ldots, b_k$.