, 2020

Let $G(x_1, \dots, x_n)$ be the function of the $n$ variables $x_1, \dots, x_n$ on the LHS of the required identity. Since both sides of the identity are rational functions, it suffices to prove it when all $x_i \notin \{-1, 1\}$. Define
$$f(t) = \prod_{i=1}^{n} (1 - x_i t),$$
and note that
$$f(x_i) = (1 - x_i^2) \prod_{j \ne i} (1 - x_i x_j).$$
Using the nodes $+1, -1, x_1, \dots, x_n$, the Lagrange interpolation formula gives us the following expression for $f$:
$$f(x) = \sum_{i=1}^{n} f(x_i) \frac{(x-1)(x+1)}{(x_i-1)(x_i+1)} \prod_{j \ne i} \frac{x-x_j}{x_i-x_j} \\ + f(1) \frac{x+1}{1+1} \prod_{i=1}^{n} \frac{x-x_i}{1-x_i} + f(-1) \frac{x-1}{-1-1} \prod_{i=1}^{n} \frac{x-x_i}{-1-x_i}.$$
Note that the coefficient of $t^{n+1}$ in $f(t)$ is zero, since $f$ has degree $n$. Thus, the coefficient of $t^{n+1}$ in the above expression of $f$ gives
$$\begin{aligned} 0 &= \sum_{i=1}^{n} \frac{f(x_i)}{\prod_{j \ne i} (x_i - x_j)(x_i - 1)(x_i + 1)} \\ &\quad + \frac{f(1)}{\prod_{j \ne i} (1 - x_j)(1 + 1)} + \frac{f(-1)}{\prod_{j \ne i} (-1 - x_j)(-1 - 1)} \\ &= -G(x_1, \dots, x_n) + \frac{1}{2} + \frac{(-1)^{n+1}}{2}. \end{aligned}$$
This implies the required identity. $\square$