Showing $\sum_{i=1}^n\prod_{j\neq i}x_j^2\prod_{q\neq p,p\neq i}x_q-x_p=(\sum_{i=1}^n\prod_{j\neq i}x_j)(\prod_{q\neq p}(x_q-x_p))$ Let $(x_i)_{i=1}^n$ be a sequence of distinct nonzero real numbers (the "distinct nonzero" assumption is likely not necessary, but we may assume it. If you can discuss the case without this assumption, I'll likely accept the answer). I'd like to show $$\sum_{i=1}^n\left(\prod_{j \neq i} x_j^2 \prod_{q \neq p, p \neq i} (x_q-x_p)\right) =\left( \sum_{i=1}^n \prod_{j \neq i} x_j\right)\prod_{q \neq p} (x_q - x_p).$$
For instance, with $n=2$ this reads $x_2^2(x_1-x_2) + x_1^2(x_2-x_1) = (x_1 + x_2)(x_1-x_2)(x_2-x_1)$, which is true: $x_2^2(x_1-x_2) + x_1^2(x_2-x_1) = (x_1^2 - x_2^2)(x_2-x_1)=(x_1+x_2)(x_1-x_2)(x_2-x_1)$. We can verify the $n=3$ case on Wolfram here.
I tried solving it with brute force algebra, but haven't had luck with that.  Of course, for $n>2$, there isn't algebraic trick as simple as difference of squares, which we can do with $n=2$.
Another idea: the problem gives Lagrange interpolation vibes. The "distinct nonzero" assumption could be useful for dividing by $x_j - x_i$ at some point, which is relevant for Lagrange interpolation. Second idea: identifying both sides as the determinant of some matrix? The expressions look not too dissimilar from the determinant of the Vandermonde matrix.
 A: Your guess about determinants is right on the money; by Jacobi's bialternant formula for the Schur polynomials, together with a known identification between a special case of the Schur polynomials and the elementary symmetric polynomials, the RHS is the Vandermonde-like determinant
$$\det \left[ \begin{array}{cccc} x_1^n & x_2^n & \cdots & x_n^n \\
 x_1^{n-1} & x_2^{n-2} & \cdots & x_n^{n-1} \\
 \vdots & \vdots & \ddots & \vdots \\ 
 x_1^2 & x_2^2 & \cdots & x_n^2 \\
 1 & 1 & \cdots & 1
 \end{array} \right].$$
Now the LHS is obtained by cofactor expansion along the last row; the $i^{th}$ term is a Vandermonde determinant for $x_j, j \neq i$ but with the columns multiplied by $x_j^2, j \neq i$, and we factor these out. Then we just have to check that the signs match.
No assumptions on the $x_i$ are necessary; this is a polynomial identity so it's true for the $x_i$ being formal variables.
Edit: Ah, there's a small fixable error here. Both sides have an extra factor of the Vandermonde determinant $\prod_{p < q} (x_p - x_q)$ that needs to be divided out first. This makes it less obvious that the signs will match up but I can't imagine that not happening.
A: The interpolation idea also works.  The polynomial of degree $n$
$$
1-\sum_{i=1}^n\frac{x}{x_i}\prod_{j\ne i}\frac{1-x/x_j}{1-x_i/x_j}
$$
has zeros $x_1, \ldots, x_n$ and constant term $1,$ so is equal to
$\prod_{i=1}^n(1 - x/x_i).$ Equating coefficients of $x$ in this
polynomial identity, we get the stated result in the form
\begin{equation}
\label{4528853:eq:1}\tag{1}
\sum_{i=1}^n\bigg(x_i\prod_{j\ne i}
\left(1 - \frac{x_i}{x_j}\right)\bigg)^{-1} \!\! =
\sum_{i=1}^n\frac1{x_i}.
\end{equation}

(Expanding the last part of the answer, as requested in a comment:)
Rewrite \eqref{4528853:eq:1} as:
$$
\left(\prod_{i=1}^nx_i\right)\Bigg(\sum_{i=1}^n
\bigg(x_i^2\prod_{j\ne i}(x_j-x_i)\bigg)^{-1}\Bigg) =
\sum_{i=1}^n\frac1{x_i}.
$$
Multiply both sides of this identity by $\prod_{i=1}^nx_i,$
obtaining:
\begin{equation}
\label{4528853:eq:2}\tag{2}
\left(\prod_{i=1}^nx_i\right)^2\Bigg(\sum_{i=1}^n
x_i^{-2}\bigg(\prod_{j\ne i}(x_j-x_i)\bigg)^{-1}\Bigg) =
\sum_{i=1}^n\prod_{j\ne i}x_j.
\end{equation}
Rewrite the left hand side of \eqref{4528853:eq:2} as:
$$
\sum_{i=1}^n\bigg(\bigg(\prod_{j\ne i}x_j^2\bigg)
\bigg(\prod_{j\ne k\ne i}(x_j-x_k)\bigg)
\bigg(\prod_{j\ne k}(x_j-x_k)\bigg)^{-1}\bigg).
$$
Multiply both sides of this version of \eqref{4528853:eq:2} by
$\prod_{j\ne k}(x_j-x_k),$ obtaining:
$$
\sum_{i=1}^n\bigg(\bigg(\prod_{j\ne i}x_j^2\bigg)
\bigg(\prod_{j\ne k\ne i}(x_j-x_k)\bigg)\bigg) =
\bigg(\sum_{i=1}^n\prod_{j\ne i}x_j\bigg)
\bigg(\prod_{j\ne k}(x_j-x_k)\bigg).
$$
The required identity is obtained by (i) omitting some parentheses,
and (ii) replacing the bound variables $j, k$ by $q, p$ respectively
in two subexpressions. $\square$
A: The question asks for a proof of a multivariable polynomial identity. The
sequence of variables $\,x_1,x_2,x_3,\dots\,$ may be more conveniently
labeled as $\,a_,b_,c_,\dots\,$ for the purpose of giving examples. In the
course of my proof I will generalize and prove other identities also.
To begin with, I define
$$ P_n(x) := \prod_{i=1}^n x_i, \quad
W_n(x) := \prod_{i<j} x_i - x_j, \quad 
V_n(x) := \prod_{i\ne j} x_i - x_j, \tag1 $$
$$ K_{n,i}(x) := \prod_{q\ne p, p\ne i} x_q-x_p, \quad
R_{n,i}(x) := \prod_{i\ne j} x_j. \tag2 $$
Now defining the sums
$$ S_{n,k}(x) := \sum_{i=1}^n R_{n,i}(x)^k K_{n,i}(x),
 \qquad Q_n(x) := \sum_{i=1}^n R_{n,i}(x), \tag3 $$
the problem is to prove that
$$ S_{n,2}(x) = Q_n(x)V_n(x) \qquad \text{ for all } \;\; n\ge 0. \tag4 $$
With the usual conventions for empty sums and products,
when $\,n=0,\,1,\,$ then this equation becomes $\,n=n\,$ which is
trivially true. Thus, the problem is only interesting when $\,n>1.\,$
The summands $\,K_{n,i}(x)\,$ of $\,S_{n,k}(x)\,$ have common
factors. I exhibit this by defining
$$ E_{n,i}(x) := \prod_{p<q, p,p\ne i,q\ne i} x_p-x_q \tag5 $$
(note the $p<q\,$ and $x_p-x_q$) and using factorization verify
(note the sign changes depending on $\,n\,$ and $\,i.\,$)
the following polynomial equation
$$ K_{n,i}(x) = (-1)^{n(n+1)/2}\, W_n(x)\,(-1)^i E_{n,i}(x). \tag6 $$
Thus it makes sense to prove results with the common factors
removed.

For $\,k=0\,$ this leads to proving
Lemma 1. If $n>1$ then
$$ 0 = \sum_{i=1}^n (-1)^i\,E_{n,i}(x). \tag7 $$
Proof: If any three of the $\,\{x_i\}\,$ are equal, then all
of the $\,E_{n,i}(x)\,$ are zero and the equation is trivially true.
Otherwise, there exists $\,x_p\,$ such that there are at least
$\,n\!-\!2\,$ distinct $\,x_q\,$ where $\,p\ne q.\,$ The sum is a
polynomial of degree $\,n\!-\!2\,$ in $\,x_p\,$ which has
$\,n\!-\!2\,$distinct roots $\,x_q\,$ since all of the
$\,(-1)^i\,E_{n,i}(x)\,$ are zero except if $\,i\!=\!p\,$ or
$\,i\!=\!q\,$ and these two terms are negatives of each other.
Thus, the polynomial must be equal to zero. Q.E.D.
Examples:

*

*$(n\!=\!2)\;\; 0 = -1 + 1.$

*$(n\!=\!3)\;\; 0 = (-b+c)+(a-c)+(-a+b).$

*$(n\!=\!4)\;\; 0 = -(b-c)(b-d)(c-d)+(a-c)(a-d)(c-d)-(a-d)(a-d)(b-d)+
(a-b)(a-c)(b-c).$
The identity for $\,n=4\,$ is listed as $\,id_{4,4,1,3a}\,$ in my
Special Algebraic Identities
list.

For $\,k=1\,$ this leads to proving
Lemma 2. If $n>0$ then
$$ (-1)^n\, W_n(x) = \sum_{i=1}^n R_{n,i}(x)\,(-1)^i\,E_{n,i}(x). \tag8 $$
Proof: Use Lemma 1 with $\,x_{n+1}=0\,$ to get
$$ \sum_{i=1}^n (-1)^i\,E_{n+1,i}(x) -(-1)^n\,E_{n+1,n+1}(x) = 0. \tag9 $$
Notice that factorizing gives
$$ E_{n+1,n+1}(x) = W_n(x), \;\;
   E_{n+1,i}(x) = R_{n,i}(x)E_{n,i}(x) \tag{10} $$
and substituting these in the previous equation proves the Lemma.
Q.E.D.
Examples:

*

*$(n\!=\!1)\;\;-1  = -1.$

*$(n\!=\!2)\;\; a-b = b\cdot-1+a\cdot1.$

*$(n\!=\!3)\;\; -(a-b)(a-c)(b-c) = bc(-b+c)+ac(a-c)+ab(-a+b).$
The identity for $\,n=3\,$ is listed as $\,id_{3,4,1,3b}.\,$ in my list.
list. The $\,n=4\,$ identity is listed as $\,id_{4,5,1,6a}.\,$
Many other identities given by the Lemmas are in my list.

Suppose that $\,\{z_1,\dots,z_n\}\,$ is any sequence of numbers.
Define the polynomial
$$ D_{n,i}(x) = \prod_{j\ne i} (x_j-x_i) \tag{11} $$
in order to express the Lagrange interpolation formula
$$ F(x_{n+1}) = \sum_{i=1}^n \frac{D_{n,n+1}(x)\,z_i}
    {D_{n,i}(x)\,(x_i-x_{n+1})} \tag{12} $$
where $\,F(x)\,$ is a polynomial of degree at most $\,n-1\,$
that satisfies $\,F(x_i) = z_i\,$ for $\,i=1,\dots,n.\,$
Apply this formula for the particular case $\,z_i = x_i^k\,$
where $\,0\le k<n\,$ to get
$$ x_{n+1}^k = \sum_{i=1}^n \frac{D_{n,n+1}(x)\,x_i^k}
    {D_{n,i}(x)\,(x_i-x_{n+1})}. \tag{13} $$
Further specialize with $\,x_{n+1} = 1\,$ to get
$$ -(-1)^n = \sum_{i=1}^n x_i^k\prod_{j\ne i}
    \frac{ 1-x_j }{x_j-x_i}. \tag{14} $$
Divide both sides by a common factor to get
$$ -(-1)^n \prod_{j=1}^n \frac1{1-x_j} = \sum_{i=1}^n 
\frac{x_i^k}{1-x_i} \prod_{j\ne i}
    \frac1{x_j-x_i}. \tag{15} $$

TO BE CONTINUED
A: Well I can show something strange but first a theorem :
Notably we can use the constraint :
$$\left(|\sum_{i=1}^n \prod_{j \neq i} x_j|\right)=0$$
The other side with absolute value is either zero or positive so it's an upper bound
I think we can do someting similar for the lower bound using Karamata's inequality :
Example in the case $n=3$ :
Suppose we have the inequality wich are simplified :
$$\left|ab\right|-\left|\left(a-b\right)\left(x-a\right)\right|<0,\left|xb\right|\left|ab\right|-\left|(x-b)(b-a)\right|\left|\left(a-b\right)\left(x-a\right)\right|<0,\left|xb\right|\left|ab\right|\left|xa\right|-\left|(a-x)(b-a)\right|\left|(x-b)(b-a)\right|\left|\left(a-b\right)\left(x-a\right)\right|<0$$
Then using the convexity of $f(x)=e^x$ and Karamata's inequality we get a lower bound .
On the other hand we have shown that it was an upper bound so it's an equality in this special case +homogeneity.
A: A two step approach. At first we simplify OPs expression. Then we work with Lagrange polynomials.
We can write OPs left-hand side as
\begin{align*}
\sum_{i=1}^n&\prod_{{j=1}\atop{j\neq i}}^{n}x_j^2
\prod_{{1\leq p,q\leq n}\atop{q\neq p, p\neq i}}\left(x_q-x_p\right)\\
&=\sum_{i=1}^n\prod_{{j=1}\atop{j\neq i}}^{n}x_j^2
\left(\prod_{{1\leq p,q\leq n}\atop{q\neq p}}\left(x_q-x_p\right)\right)
\left(\prod_{{1\leq q\leq n}\atop{q\neq i}}\left(x_q-x_i\right)\right)^{-1}\\
&=\sum_{i=1}^n\prod_{{j=1}\atop{j\neq i}}^{n}\frac{x_j^2}{x_j-x_i}\left(\prod_{{1\leq p,q\leq n}\atop{q\neq p}}\left(x_q-x_p\right)\right)\tag{1.1}
\end{align*}

We can now divide OPs identity by the right-most factor from (1.1) and OPs expression boils down to
\begin{align*}
\sum_{i=1}^n\prod_{j\neq i}\frac{x_j^2}{x_j-x_i}
&=\sum_{i=1}^n\prod_{j\neq i}x_j\\\
\mathrm{resp.}\\
\color{blue}{\sum_{i=1}^n\frac{1}{x_i}\prod_{j\neq i}\frac{x_j}{x_j-x_i}}
&\color{blue}{=\sum_{j=1}^n\frac{1}{x_j}\qquad\qquad n\geq 1\quad\mathrm{integer}}\tag{1.2}
\end{align*}

The left-hand side of (1.2) is just the Lagrange polynomial $L(x)$
\begin{align*}
L(x)=\sum_{i=1}^n\frac{1}{x_i}\prod_{{j=1}\atop{j\ne i}}^n\frac{x_j-x}{x_i-x_j}\tag{2.1}
\end{align*}
evaluated at $0$ with
\begin{align*}
L(x_i)=\frac{1}{x_i}\qquad\qquad 1\leq i\leq n\tag{2.2}
\end{align*}
We derive from (2.1) and (2.2) a product representation
\begin{align*}
\color{blue}{1-xL(x) =1-x\sum_{i=1}^n\frac{1}{x_i}\prod_{{j=1}\atop{j\ne i}}^n\frac{x_j-x}{x_i-x_j}= \prod_{j=1}^n\left(1-\frac{x}{x_j}\right)}\tag{3}
\end{align*}
since $1-x_jL\left(x_j\right)=0, 1\leq j\leq n$.

Denoting with $[n]=\{1,2,\ldots,n\}$ we derive from (3) by subtracting $1$ and division by $-x$:
\begin{align*}
\color{blue}{\sum_{i=1}^n}
\color{blue}{\frac{1}{x_i}\prod_{{j=1}\atop{j\ne i}}^n\frac{x_i}{x_j-x_i}}
&=-\lim_{x\to 0}\frac{1}{x}\sum_{k=1}^n\sum_{{S\subseteq [n]}\atop{|S|=k}}
\prod_{j\in S}\left(-\frac{x}{x_j}\right)\tag{4.1}\\
&=-\sum_{{S\subseteq [n]}\atop{|S|=1}}\prod_{j\in S}\left(-\frac{1}{x_j}\right)\tag{4.2}\\
&\,\,\color{blue}{=\sum_{j=1}^n\frac{1}{x_j}}\tag{4.3}
\end{align*}
and the claim (1) follows.

Comment:

*

*In (4.1) we expand the product of the right-hand side of (3). We start with $k=1$ since $1$ has already been subtracted.


*In (4.2) we note the summands with $k>1$ evaluate to zero by taking the limit, leaving the summand with $k=1$.


*In (4.3) we can make a final simplification, since we sum over all $1$-element subsets of $[n]$.
