22nd ASU 1988 Problem 18 Find the minimum value of xy/z + yz/x + zx/y for positive reals x, y, z with x 2 + y 2 + z 2 = 1. Solution
: min √3 when all equal. Let us consider z to be fixed and focus on x and y. Put f(x, y, z) = xy/z + yz/x + zx/y. We have f(x, y, z) = p/z + z(1-z 2 )/p = (p + k 2 /p)/z, where p = xy, and k = z√(1-z 2 ). Now p can take any value in the range 0  √3. If z > 1/√5, then the minimum of f(x, y, z) occurs at x = y and is x 2 /z + z + z = (1-z 2 )/(2z) + 2z = 3z/2 + 1/(2z) = (√3)/2 (z√3 + 1/(z√3) ≥ √3 with equality at z = 1/√3 (and hence x = y = 1/√3 also). 22nd ASU 1988 © John Scholes jscholes@kalva.demon.co.uk 26 June 2002