Prove that matrix is positive definite

Update: I originally claimed to prove that $A$ is strictly positive definite, but there was a bug in the strictness part. I have revised the proof to show that $A$ is positive semidefinite. For an example to see that $A$ need not be strictly positive definite let $x_i=y_i$ for all $i$. Then $A = xx^T$ is rank one.

---

For any sequence $z = (z_1,\ldots, z_n)$ of nonnegative numbers, the matrix $B(z)$ with entries $[B(z)]_{ij} = \min(z_i, z_j)$ is positive semidefinite. Given this, we set $z_i = y_i/x_i$ and obtain that $A=\operatorname{diag}(x)B(z)\operatorname{diag}(x)$ is positive semidefinite.

To see that $B(z)$ is positive semidefinite note that reordering $z$ just permutes corresponding rows and columns, so assume WLOG that $z$ is sorted in nondecreasing order. Let $w_1 = z_1$ and $w_i = z_i - z_{i-1}$ for $i>1$. Let $J$ be the matrix with ones on the upper triangle (including the diagonal) and zeros below. Then $w\geq 0$ so $B(z) = J^T\operatorname{diag}(w)J$ is positive semidefinite.

This is a very similar proof but based on $\mathbb{E}B_{z_{i}}B_{z_{j}}=\min(z_{i},z_{j})$ where $B_{z}$ is the standard brownian motion starting at zero:

$ 0\leq \mathbb{E}\left(\sum_{j} a_{j} x_{j} B_{\frac{y_{j}}{x_{j}}} \right)^{2}=\sum_{i,j}a_{i}a_{j}\min(x_{i}y_{j},x_{j}y_{i}) $