Convexity of difference of log-sum-exp: $f(x_1, x_2, x_3, x_4) = \log(e^{x_1} + e^{x_2}) - \log(e^{x_1} + e^{x_2} + e^{x_3} + e^{x_4})$

(This answer builds up on the first comment to the question above).

Short answer: no, the function is not concave.

Instead of analyzing the full $4 \times 4$ hessian, we can start by restricting our attention to a subspace of the input, e.g., the line induced by setting $x_2 = x_3 = x_4 = 0$. Along this line, the original function can be rewritten as a univariate function $\tilde{f}(x) = \log(e^x + 1) - \log(e^x + 3)$.

the second derivative of $\tilde{f}$ is $$ \frac{d^2\tilde{f}}{dx^2} = \frac{e^x(6 - 2 e^{2x})}{(e^x + 1)^2 (e^x + 3)^2} $$ it is easy to see that the second derivative is positive for $x = 0$, hence $\tilde{f}$ is not concave (furthemore, it is negative for $x = 1$, so it is not convex either). We conclude that $f$ is not concave.

Note that the type of functions described here (difference of log-sum-exps) appears in the log-likelihood function of certain statistical models of paired comparisons, such as Elimination by Aspects and team comparisons.