Let $a_1,\ldots,a_p$ be positive real numbers. Let $A = (a_{ij})_{p\times p}$ where $a_{ij} = 1/(a_i+a_j)$. How to prove that
$$\det A = 2^{-p}\prod_{j=1}^p \frac{1}{a_j}\prod_{1\leq j<k\leq p}
\biggl(\frac{a_j-a_k}{a_j+a_k}\biggr)^2.$$
The case when $p=2$ is obvious and I am stuck in the inductive procession.
Best Answer
Note that \begin{align} &D_{n} = \det\left(\dfrac{1}{a_{i} + b_{j}}\right)_{n \times n} \\ =& \begin{vmatrix} \dfrac{1}{a_{1}+b_{1}} & \dfrac{1}{a_{1}+b_{2}} & \cdots & \dfrac{1}{a_{1}+b_{n}} \\ \dfrac{1}{a_{2}+b_{1}} & \dfrac{1}{a_{2}+b_{2}} & \cdots & \dfrac{1}{a_{2}+b_{n}} \\ \vdots & \vdots & \ldots & \vdots \\ \dfrac{1}{a_{n}+b_{1}} & \dfrac{1}{a_{n}+b_{2}} & \cdots & \dfrac{1}{a_{n}+b_{n}} \\ \end{vmatrix} \\ =& \begin{vmatrix} \dfrac{1}{a_{1}+b_{1}} - \dfrac{1}{a_{n}+b_{1}} & \dfrac{1}{a_{1}+b_{2}} - \dfrac{1}{a_{n}+b_{2}} & \cdots & \dfrac{1}{a_{1}+b_{n}} - \dfrac{1}{a_{n}+b_{n}} \\ \dfrac{1}{a_{2}+b_{1}} - \dfrac{1}{a_{n}+b_{1}} & \dfrac{1}{a_{2}+b_{2}} - \dfrac{1}{a_{n}+b_{2}} & \cdots & \dfrac{1}{a_{2}+b_{n}} - \dfrac{1}{a_{n}+b_{n}} \\ \vdots & \vdots & \ldots & \vdots \\ \dfrac{1}{a_{n}+b_{1}} & \dfrac{1}{a_{n}+b_{2}} & \cdots & \dfrac{1}{a_{n}+b_{n}} \\ \end{vmatrix} \\ =& \begin{vmatrix} \dfrac{a_{n}-a_{1}}{(a_{1}+b_{1})(a_{n}+b_{1})} & \dfrac{a_{n}-a_{1}}{(a_{1}+b_{2})(a_{n}+b_{2})} & \cdots & \dfrac{a_{n}-a_{1}}{(a_{1}+b_{n})(a_{n}+b_{n})} \\ \dfrac{a_{n}-a_{2}}{(a_{2}+b_{1})(a_{n}+b_{1})} & \dfrac{a_{n}-a_{2}}{(a_{2}+b_{2})(a_{n}+b_{2})} & \cdots & \dfrac{a_{n}-a_{2}}{(a_{2}+b_{n})(a_{n}+b_{n})} \\ \vdots & \vdots & \ldots & \vdots \\ \dfrac{1}{a_{n}+b_{1}} & \dfrac{1}{a_{n}+b_{2}} & \cdots & \dfrac{1}{a_{n}+b_{n}} \\ \end{vmatrix} \\ =& \prod_{j = 1}^{n}\dfrac{1}{a_{n}+b_{j}}\begin{vmatrix} \dfrac{a_{n}-a_{1}}{(a_{1}+b_{1})} & \dfrac{a_{n}-a_{1}}{(a_{1}+b_{2})} & \cdots & \dfrac{a_{n}-a_{1}}{(a_{1}+b_{n})} \\ \dfrac{a_{n}-a_{2}}{(a_{2}+b_{1})} & \dfrac{a_{n}-a_{2}}{(a_{2}+b_{2})} & \cdots & \dfrac{a_{n}-a_{2}}{(a_{2}+b_{n})} \\ \vdots & \vdots & \ldots & \vdots \\ 1 & 1 & \cdots & 1 \\ \end{vmatrix} \\ =& \dfrac{\prod_{1 \leqslant i < n}(a_{n} - a_{i})}{\prod_{j = 1}^{n}(a_{n}+b_{j})}\begin{vmatrix} \dfrac{1}{(a_{1}+b_{1})} & \dfrac{1}{(a_{1}+b_{2})} & \cdots & \dfrac{1}{(a_{1}+b_{n})} \\ \dfrac{1}{(a_{2}+b_{1})} & \dfrac{1}{(a_{2}+b_{2})} & \cdots & \dfrac{1}{(a_{2}+b_{n})} \\ \vdots & \vdots & \ldots & \vdots \\ 1 & 1 & \cdots & 1 \\ \end{vmatrix} \\ \end{align} and \begin{align} & \begin{vmatrix} \dfrac{1}{(a_{1}+b_{1})} & \dfrac{1}{(a_{1}+b_{2})} & \cdots & \dfrac{1}{(a_{1}+b_{n})} \\ \dfrac{1}{(a_{2}+b_{1})} & \dfrac{1}{(a_{2}+b_{2})} & \cdots & \dfrac{1}{(a_{2}+b_{n})} \\ \vdots & \vdots & \ldots & \vdots \\ 1 & 1 & \cdots & 1 \\ \end{vmatrix} \\ =& \begin{vmatrix} \dfrac{1}{(a_{1}+b_{1})} - \dfrac{1}{(a_{1}+b_{n})} & \dfrac{1}{(a_{1}+b_{2})} - \dfrac{1}{(a_{1}+b_{n})} & \cdots & \dfrac{1}{(a_{1}+b_{n})} \\ \dfrac{1}{(a_{2}+b_{1})} - \dfrac{1}{(a_{2}+b_{n})} & \dfrac{1}{(a_{2}+b_{2})} - \dfrac{1}{(a_{2}+b_{n})} & \cdots & \dfrac{1}{(a_{2}+b_{n})} \\ \vdots & \vdots & \ldots & \vdots \\ 1 - 1 & 1 - 1 & \cdots & 1 \\ \end{vmatrix} \\ =& \begin{vmatrix} \dfrac{b_{n}-b_{1}}{(a_{1}+b_{1})(a_{1}+b_{n})} & \dfrac{b_{n}-b_{2}}{(a_{1}+b_{2})(a_{1}+b_{n})} & \cdots & \dfrac{1}{(a_{1}+b_{n})} \\ \dfrac{b_{n}-b_{1}}{(a_{2}+b_{1})(a_{2}+b_{n})} & \dfrac{b_{n}-b_{2}}{(a_{2}+b_{2})(a_{2}+b_{n})} & \cdots & \dfrac{1}{(a_{2}+b_{n})} \\ \vdots & \vdots & \ldots & \vdots \\ 0 & 0 & \cdots & 1 \\ \end{vmatrix} \\ =& \dfrac{\prod_{1 \leqslant j < n}(b_{n} - b_{j})}{\prod_{1 \leqslant i < n}(a_{i}+b_{n})}\begin{vmatrix} \dfrac{1}{a_{1}+b_{1}} & \dfrac{1}{a_{1}+b_{2}} & \cdots & 1 \\ \dfrac{1}{a_{2}+b_{1}} & \dfrac{1}{a_{2}+b_{2}} & \cdots & 1 \\ \vdots & \vdots & \ldots & \vdots \\ 0 & 0 & \cdots & 1 \\ \end{vmatrix} \\ =& \dfrac{\prod_{1 \leqslant j < n}(b_{n} - b_{j})}{\prod_{1 \leqslant i < n}(a_{i}+b_{n})}D_{n-1} \end{align} then we have $$D_{n} = \dfrac{\prod_{1 \leqslant i < n}(a_{n} - a_{i})}{\prod_{j=1}^{n}(a_{n}+b_{j})} \cdot \dfrac{\prod_{1 \leqslant j < n}(b_{n} - b_{j})}{\prod_{1 \leqslant i < n}(a_{i}+b_{n})}D_{n-1} = \dfrac{\prod_{1 \leqslant i < n}(a_{n}-a_{i})(b_{n}-b_{i})}{(a_{n}+b_{n})\prod_{1 \leqslant j < n}(a_{n}+b_{j})\prod_{1 \leqslant i < n}(a_{i}+b_{n})}D_{n-1}$$ recall that $D_{0} = 1$, hence we have \begin{align} D_{n} =& \dfrac{\prod_{j = 1}^{n}\prod_{1 \leqslant i < j}(a_{j}-a_{i})(b_{j}-b_{i})}{\prod_{1 \leqslant i = j \leqslant n}(a_{i}+b_{j})\prod_{i=1}^{n}\prod_{1 \leqslant j < i}(a_{i}+b_{j})\prod_{j=1}^{n}\prod_{1 \leqslant i < j}(a_{i}+b_{j})} \\ =& \dfrac{\prod_{1 \leqslant i < j \leqslant n}(a_{j}-a_{i})(b_{j}-b_{i})}{\prod_{\substack{1 \leqslant i \leqslant n \\ 1 \leqslant j \leqslant n}}(a_{i}+b_{j})} \\ \end{align}
The penultimate step is the formula when $b = a$.