Proof. Since $R$ is commutative, we can apply the binomial formula.We start out with \begin{equation} \left(\sum_{i=1}^n x_iy_i\right)^2 =\sum_{i=1}^n (x_i^2y_i^2) +\sum_{1 \leq i< j\leq n} 2 x_iy_jx_jy_i \end{equation}Using the binomial formula, we see that$$(x_iy_j -x_jy_i)^2 =x_i^2y_j^2 -2x_ix_jy_iy_j +x_j^2y_i^2$$ So we get

\begin{eqnarray} \left(\sum\limits_{i=1}^n x_iy_i\right)^2 +\sum\limits_{1 \leq i< j\leq n}^n(x_iy_j -x_jy_i)^2 &=&\sum\limits_{i=1}^n \left(x_i^2y_i^2\right) +\sum\limits_{1 \leq i< j\leq n}^n \left(x_i^2y_j^2 +x_j^2y_i^2\right) \\ &=&\left(\sum\limits_{i=1}^n x_i^2\right)\left(\sum\limits_{i=1}^n y_i^2\right) \end{eqnarray}Note that changing the roles of $i$ and $j$ in $x_iy_j -x_jy_i$ , we get$$x_jy_i -x_iy_j =-(x_iy_j -x_jy_i)$$ but the negative sign will disappear when we square. So we can rewrite the last equation to \begin{equation} \left(\sum\limits_{i=1}^n x_iy_i\right)^2 +\sum\limits_{1 \le i <j \le n} (x_iy_j -x_jy_i)^2 =\left(\sum_{i=1}^n x_i^2\right)\left(\sum_{i=1}^n y_i^2\right). \end{equation}This is equivalent to the stated identity.