Showing that $(A_{ij})=\left(\frac1{1+x_i+x_j}\right)$ is positive semidefinite

Question

Consider the matrix $A$ whose coefficients are $A_{ij} = \frac{1}{1+x_i+x_j} $ where we have $ x_i \geq 0$ and $ x_j \geq 0$ for $ i,j=1,2,\dots,n$. How can I prove that this matrix is positive semidefinite for arbitrary $n$? Using first principal minors, I proved that this is positive semi-definite for $n=2$ but I could not generalize it.

Answer 1

Edit: I first foolishly proved that $A$ is positive definite in general. But that's obviously wrong, as $x_1=\ldots=x_n=0$ yields a rank $1$ matrix. So I removed the wrong argument.

Consider the matrix $$ A_t=\left(t^{x_i+x_j}\right)_{i,j} $$ and show this is a continuous path of symmetric positive semidefinite for $t\in[0,1]$.

To see this, introduce the matrix $$ B_t=\left(\frac{t^{x_j}}{\sqrt{n}}\right)_{i,j} $$ and observe that $$ A_t=B_t^*B_t. $$

Then integrate between $0$ and $1$, you find $$ \int_0^1A_tdt=A $$ where integration is performed componentwise.

Note that $A$ is obviously symmetric.

Now fix a vector $X$. Since $(X,A_tX)\geq 0$ for all $t\in[0,1]$, the integration of this continuous function yields $$\int_0^1(X,A_tX)dt=(X,AX)\geq 0$$ for all $X$.

So $A$ is positive semidefinite.