At the awards ceremony, Jacob Tsimerman, one of the Canadian winners, presented the rest of us with a stronger form of Schur that he'd just received in an email. It runs thus:
a3 + b3 + c3 + 3abc ≥ ab√(2a2 + 2b2) + bc√(2b2 + 2c2) + ca√(2c2 + 2a2)
and for most of MOP, nobody could prove it (although it was verified by TI-89). We kept on writing it on the board before lectures, and on the last night of MOP, Kiran came up a solution, but he needed Maple to take care of the messy algebra.
So here's the challenge: find a reasonably nice solution. Any ideas?