16th Irish 2003 Problem B4 Given real positive a, b, find the largest real c such that c ≤ max(ax + 1/(ax), bx + 1/bx) for all positive real x.
1/√(ab) Solution The curve ax + 1/(ax) has a minimum at x = 1/a (easily checked by differentiation or completing the square), so if a = b, the answer is obviously c = 1/a. If a and b are unequal, then the situation is as shown in the graph and the required c is the value of x at which the curves intersect. That is ax + 1/ax = bx + 1/bx, or (a - b)x = 1/bx - 1/ax = (a - b)/abx. Hence x = 1/abx, or x 2 = 1/ab. 16th Irish 2003 © John Scholes jscholes@kalva.demon.co.uk 14 June 2003 Last updated/corrected 14 Jun 03