Given parameters $w_1, \ldots, w_n > 0$ with $\sum_{j=1}^n w_j = 1$ ( $n\ge 2$), define function
$$ f(x_1, \ldots, x_{n-1})= \sum_{j=1}^{n} \frac{x_j}{1-x_j} \cdot \sum_{j=1}^n \frac{w_j}{1-w_j} - \sum_{j=1}^n \frac{x_j}{1-w_j} \cdot \sum_{j=1}^n \frac{w_j}{1-x_j} $$
with $$x_n = 1-\sum_{j=1}^{n-1}x_j$$ for $x_1, \dots, x_{n-1} \in (0,1)$.
For example, for $n=3$ the function is
$$ f(x_1, x_2)= \left ( \frac{x_1}{1-x_1} + \frac{x_2}{1-x_2}+ \frac{1- x_1-x_2}{x_1+x_2}\right) \left ( \frac{w_1}{1-w_1} + \frac{w_2}{1-w_2}+ \frac{w_3}{1-w_3}\right)- \left ( \frac{w_1}{1-x_1} + \frac{w_2}{1-x_2}+ \frac{w_3}{x_1+x_2}\right) \left ( \frac{x_1}{1-w_1} + \frac{x_2}{1-w_2}+ \frac{1- x_1-x_2}{1-w_3}\right)$$
Working on a conjecture in this MO question, for $n= 3, 4, 5$, I want to numerically check whether the Hessian matrix of $f(x_1, \ldots, x_{n-1})$ is positive semi-definite over the hypercube $(0,1)^{n-1}$ for any parameters $w_1, \ldots, w_n > 0$ with $\sum_{j=1}^n w_j = 1$.
I could manage the case $n=3$ using a desmos graph, which shows the answer is yes. Could you help me efficiently use Mathematica to handle the two remaining cases $n=4,5$?
I found a related question discussing how one can check whether a matrix is positive semi-definite by Mathematica.