22nd BMO 1986 Problem 2 Find the maximum value of x 2 y + y 2 z + z 2 x for reals x, y, z with sum zero and sum of squares 6.
6 (eg x = 2 cos 40 o , y = 2 cos 80 o , z = 2 cos 160 o ). Solution Put u = x 2 y+y 2 z+z 2 x, v = xy 2 +yz 2 +zx 2 , and w = xyz. We have 0 = (x+y+z) 2 = x 2 +y 2 +z 2 +2(xy+yz+zx), so xy+yz+zx = -3. Squaring, 9 = x 2 y 2 +y 2 z 2 +z 2 x 2 +2(x 2 yz+xy 2 z+xyz 2 ) = x 2 y 2 +y 2 z 2 +z 2 x 2 + 2xyz(x+y+z) = x 2 y 2 +y 2 z 2 +z 2 x 2 . We have x 3 +y 3 +z 3 = (x+y+z)(x 2 +y 2 +z 2 -xy-yz-zx) + 3xyz = 3w. Hence also x 3 y 3 +y 3 z 3 +z 3 x 3 = (xy+yz+zx)(x 2 y 2 +y 2 z 2 +z 2 x 2 -xy 2 z-xyz 2 -x 2 yz) + 3x 2 y 2 z 2 = (xy+yz+zx)(x 2 y 2 +y 2 z 2 +z 2 x 2 ) + 3w 2 = -3·9 + 3w 2 = -27 + 3w 2 . Also (x+y+z)(xy+yz+zx) = u+v+3w, so u+v = -3w. Multiplying out, we find uv = (x 3 y 3 +y 3 z 3 +z 3 x 3 ) + 3x 2 y 2 z 2 + xyz(x 3 +y 3 +z 3 ) = -27 + 3w 2 + 3w 2 + 3w 2 = 9w 2 - 27. Hence u is a root of the quadratic u 2 + 3wu + (9w 2 -27) = 0. Completing the square in w (not u) we get (3w + u/2) 2 + 3u 2 /4 - 27 = 0, so 3u 2 /4 - 27 ≤ 0 and hence u ≤ 6. For equality we need w = -1. Remembering that x+y+z = 0, xy+yz+zx = -3, x, y, z are the roots of k 3 - 3k + 1 = 0. If we put k = 2 cos θ, then 8 cos 3 θ - 6 cos θ + 1 = 0. But cos 3θ = 4 cos 3 θ - 3 cos θ, so cos 3θ = -1/2. Hence 3θ = 2π/3, 4π/3 or 8π/3, so x, y, z = 2 cos 2π/9, 2 cos 4π/9, 2 cos 8π/9. Comment. This is all fairly unmotivated. Does anyone have a more straightforward solution? 22nd BMO 1986 © John Scholes jscholes@kalva.demon.co.uk 11 February 2004 Last corrected/updated 11 Feb 04