Let $\mu$ be a probability measure on $[0,\infty)$ and $X_1, \dots, X_4 \sim \mu$ independent. Then what can be said about the probability that $X_1 + X_2 + X_3 < 2 X_4$?

More precisely, what is the value of $$\tag{1}\sup_\mu \mathbf{P}[X_1 + X_2 + X_3 < 2 X_4] \quad ?$$ Here's what I know:

• Numerically, taking $\mu$ to be a Gamma distribution at parameter $\alpha \approx 0.5$ gives a value of $\approx 0.3080$ (clearly independently of scale). This is the largest value I've found among the distributions I've tried.
• In the other direction, a symmetrization argument shows that $$\sup_\mu \mathbf{P}[X_1 + X_2 + X_3 < 3 X_4] = \frac{1}{2},$$ and this serves as an easy upper bound on (1).
• There are several similar problems which I've been able to solve: $$\sup_\mu \mathbf{P}[X_1 + X_2 < X_3] = \frac{1}{3}, \qquad \sup_\mu \mathbf{P}[X_1 + X_2 + X_3 < X_4 + X_5] = \frac{2}{5}.$$ In both of these cases, the supremum is approached by a sequence of families of distributions that are uniform on a large finite set with exponential spacing, so that $+$ effectively becomes $\max$. But such distributions do not approach the supremum in (1).