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 gives an upper bound of $\frac{1}{2}$ on (1). Indeed putting each of the four variables on the right-hand side gives four versions of the inequality $X_1 + X_2 + X_3 < 2 X_4$, all with the same probability. At most two of these can be satisfied jointly (since if three of them held jointly, then adding them produces a contradiction). Then the upper bound of $\frac{1}{2}$ follows by exchangeability: if four symmetric events are such that at most two can happen jointly, then the probability of each is at most $\frac{1}{2}$. (Think of the induced distribution on $\{0,1\}^4$, which must be supported on sequences with at most two $1$'s.)
• 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 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).