[[Please see SECOND EDIT at the bottom before bothering to read this]]
I'm not sure that $B^3=\{(a_1,a_2,a_3)|2a_2≤a_1+1, a_3<1\}$ is the right answer. I find that the feasible set for $n=3$ includes a region defined by
$${3a_2a_3 + a_2 - a_3 -1 \over 3 + a_2 + a_3 - a_2a_3} < a_1$$
(following the convention $a_1 < a_2 < a_3$). The denominator on the left hand side is always positive, but the numerator is negative when $a_2=1/2$, which means there are feasible configurations with $2a_2 > a_1+1$ --- just pick a value of $a_2$ slightly larger than $1/2$ that keeps the left hand side negative and then pick $a_1$ smaller than $2a_2-1$.
Note, I'm not saying that this inequality defines the feasible set, merely that it specifies a portion of it.
Here's my reasoning. The motorcyclist will first encounter walker 3, and can transport him back to walker 2. He can then effectively "bind" those two walkers together, in a way I'll describe below, and slow them down to a speed arbitrarily close to the left-hand side above, while walker 1 saunters by. When walker 1 has a sufficient lead, the motorcyclist can "release" walker 2 and either "hold" walker 3 in place or transport him some distance back before releasing him -- all timed so that walkers 2 and 3 catch up with 1 at the finish line.
Here's what I mean by "binding" walkers 2 and 3 together. For convenience, let's use a number line that has walker 3 (being carried left on the motorcycle) meet walker 2 at $x=0$. For a short time $\delta$, let walker 3 continue heading left on the motorcycle, so that he's now at $x_3 = -\delta$ while walker 2 is at $x_2 = \delta a_2$. Now drop off walker 3 and zip off to catch walker 2. This will take time $\Delta = \delta(1+a_2)/(1-a_2)$, at which time walker 3 is at $\Delta a_3-\delta$ and walker 2 is at $(\Delta + \delta)a_2$. If the mortorcyclist now picks up walker 2 and heads back to the left, they meet walker 3 after time $$\Delta' = {(\Delta + \delta)a_2 - (\Delta a_3 - \delta) \over 1+a_3}$$ at which point they are at $$x = \Delta a_3 - \delta + \Delta'a_3$$ and the process can repeat (i.e., take walker 3 back for time $\delta$, etc.). By taking $\delta$ small, the two walkers can be kept close together, and their average speed over a complete cycle is
$${\Delta a_3 - \delta + \Delta'a_3 \over \delta + \Delta + \Delta'} = {3a_2a_3 + a_2 - a_3 -1 \over 3 + a_2 + a_3 - a_2a_3}$$
(Caveat: I did the algebra here three times by hand, and got the same answer on the last two tries. You can judge for yourself the probability of the result being correct.)
It's worth noting why the inequality doesn't define the feasible set. It's because even if the inequality is (slightly) violated, there might be a $\delta$ for which the motorcyclist encounters walker 1 while he's transporting walker 3 to the left. If that happens, he can immediately drop off walker 3, pick up walker 1, and zip him some distance to the right, and then come back for walkers 2 and/or 3 and position them so that everyone arrives simultaneously at the finish line.
EDIT: Forget a fair amount of what I said above. I forgot to consider that walker 1 may have a considerable distance to make up while walkers 2 and 3 are bound together. The safest thing to say is that the feasible set includes a region defined by
$${3a_2a_3 + a_2 - a_3 -1 \over 3 + a_2 + a_3 - a_2a_3} < 0$$
which describes a setting where the bound pair move to the left while walker 1 proceeds to the right.
SECOND EDIT: I should have tried doing the algebra one more time by hand before posting this. The correct average speed (I think) is
$${\Delta a_3 - \delta + \Delta'a_3 \over \delta + \Delta + \Delta'} = {3a_2a_3 + a_2 + a_3 -1 \over 3 + a_2 + a_3 - a_2a_3}$$
Note the $+a_3$ is the numerator, which corrects the $-a_3$ I originally obtained. This pretty well invalidates everything I had hoped to show.