32nd BMO 1996 Problem 4 Find all positive real solutions to w + x + y + z = 12, wxyz = wx + wy + wz + xy + xz + yz + 27.
w = x = y = z = 3 is the only solution. Solution Put p = √(wxyz). By AM/GM applied to wx, wy, wz, xy, xz, yz we have p ≤ (wx + wy + wz + xy + xz + yz)/6 = p 2 /6 - 27/6, so (p - 3) 2 ≥ 36. Hence |p - 3| ≥ 6. But p is positive, so p ≥ 9. Hence wxyz ≥ 81. But applying AM/GM to w, x, y, z we have √p ≤ 3, with equality iff w = x = y = z. So wxyz ≤ 81 with equality iff w = x = y = z. 32nd BMO 1996 © John Scholes jscholes@kalva.demon.co.uk 14 Aug 2002 Last corrected/updated 22 Aug 03