When $a_1,...,a_n$ are nonnegative real numbers and $A=z_{ij}$ and $A$ is n x n where $z_{ij} = \frac{1}{1+a_i+a_j}$, find $\det(A)$.
My only observation is that the $A$ is symmetric, but that doesn't real help with computing the determinant. And because the matrix is $n$x$n$, the cofactor expansion doesn't seem plausible.
Hint. One may recognize a Cauchy determinant (a proof), $$ \begin{vmatrix} \dfrac 1 {x_1 + y_1} & \dfrac 1 {x_1 + y_2} & \cdots & \dfrac 1 {x_1 + y_n} \\ \dfrac 1 {x_2 + y_1} & \dfrac 1 {x_2 + y_2} & \cdots & \dfrac 1 {x_2 + y_n} \\ \vdots & \vdots & \ddots & \vdots \\ \dfrac 1 {x_n + y_1} & \dfrac 1 {x_n + y_2} & \cdots & \dfrac 1 {x_n + y_n} \\ \end{vmatrix}=\dfrac {\displaystyle \prod_{1 \mathop \le i \mathop < j \mathop \le n} \left({x_j - x_i}\right) \left({y_j - y_i}\right)} {\displaystyle \prod_{1 \mathop \le i, \, j \mathop \le n} \left({x_i + y_j}\right)} $$ applying it with $$ x_i=1+a_i,\quad y_j=a_j. $$