Originally Posted by

**godelproof** Hi Sudharaka my friend!

If m is very large, it can ignore A3 and first carry A1 forwards, drop him before A2 and A3, then turn back to catch A2 and carry A2 forwards, and drop him before A3 but behind A1, and turn back to carry A3 to Y when A1 and A2 just arrive at Y, too. So it is NOT A MUST that motorist first carries A3 backwards on his way to meet A1. **What the conjecture asserts, is this: A feasible plan (whether M carries A3 backwards or not) exists $\displaystyle \Longleftrightarrow$ the speeds of A1, A2 and M are such that if M goes straight to A1 and pick A1 up to catch A2, he's able to catch A2 before or just when A2 arrives at Y $\displaystyle \Longleftrightarrow$ $\displaystyle 2{a}_{2}\leq{a}_{1}+m$**. (So speed of A3 is irrelevant)

No. There you just showed that this particular plan won't work. Consider another plan: the motorist meet and carry A3 backwards and drop him a little behind A2, go forwards to pick up A2 and carry him backwards a little behind A3, and go forwards to pick up A3 and carry him a little behind A2, so on and so forth, meanwhile expecting A1 to catch up? Is this and countless other plans feasible, iff the red lettered plan in quotation is feasible? This is what the conjecture asserts!

Anyway I've proved it. The argument is long and involves many subtleties (for example, one is if you can catch A2 **before** he reaches Y as the conjecture requires, how do you know you can then make the three arrive at Y simultaneously?).

For cases involving 4 and more people, I have made no progress. I guess either they ARE indeed too complicated to handle, or require some really deep insights.