36th BMO 2000 Problem 2 Find the smallest value of x 2 + 4xy + 4y 2 + 2z 2 for positive reals x, y, z with product 32. Solution
: 96. We have (x + 2y) 2 ≥ 8xy with equality iff x = 2y. Hence (x + 2y) 2 + 2z 2 ≥ 256/z + 2z 2 with equality iff x = 2y. We have z 2 + 128/z - 48 = (z - 4) 2 (z + 8)/z ≥ 0 with equality iff z = 4 (for positive z). Hence 256/z + 2z 2 ≥ 96 with equality iff z = 4. So the expression given has minimum value 96, achieved only at x = 4, y = 2, z = 4. 36th BMO 2000 © John Scholes jscholes@kalva.demon.co.uk 18 Jul 2002