Show a specially defined matrix is positive definite 
Let $E_1, ..., E_n$ be non empty finite sets. Show that the matrix $A = (A_{ij})_{1 \leq i, j \leq n}$ defined by $A_{ij} = \dfrac{|E_i \cap E_j|}{|E_i \cup E_j|}$, is positive semi-definite.

This is part 5 of Exercise 1 in http://members.cbio.mines-paristech.fr/~jvert/svn/kernelcourse/homework/2019mva/hw.pdf .
I start with the definition but it doesn't seem very promising. Any hint, please?
 A: Assume the sets $E_i$ are all non-empty.  By deleting  rows and columns of $A$ if needed one can reduce to the case where the $E_i$ are all
 distinct.
Write $$A_{ij}= \frac{|E_i\cap E_j|}{|E_i\cup E_j|} = \frac{|E_i\cap E_j|}{|E_i|+|E_j|-|E_i\cap E_j|} =  \frac{e_{ij}}{e_i+e_j-e_{ij}} $$ where
 $e_i$ is shorthand for $|E_i|$ and so on.
Then $$A_{ij}= \frac{e_{ij}}{e_i+e_j} \left( 1 + r_{ij} + r_{ij}^2 + r_{ij}^3 + \cdots\right)\tag{*}$$ where $r_{ij}= e_{ij}/(e_i+e_j).$ The distinctness hypothesis implies $|r_{ij}|<1$ and so the convergence of the geometric series in (*).
The matrix with entries $e_{ij}$ is a Gram matrix and hence positive semidefinite.  The matrix with entries $1/(e_i+e_j) = \int_0^\infty e^
{-x|E_i|} e^{-x|E_j|} \,dx$ is also a Gram matrix so it too is psd.  Schur's product theorem tells us that their elementwise product $r_{ij}$ is thus also psd.  By Schur again, all the summand matrices in (*) are psd.  So the sum, $A$, is psd.
Examples (where all the $E_i$ are equal, so all $A_{ij}=1$, say) show that positive semi definite cannot be replaced by positive definite as in an earlier form of the  problem statement.  I suppose that strict positive definiteness holds in the reduced case, but I don't see an argument.  Numerical experiments show that the matrix $(r_{ij})$ is not strictly positive definite when (for example) the $E_i$ are all the 2-element subsets of $[4]$.  The same experiments however do show $A$ strictly positive definite.
