52nd Putnam 1991 Problem B6 Let a and b be positive numbers. Find the largest number c, in terms of a and b, such that for all x with 0 < |x| ≤ c and for all α with 0 < α < 1, we have: a α b 1-α ≤ a sinh αx/sinh x + b sinh x(1 - α)/sinh x. Solution
: c = |ln a/b|. Moderately hard. Put β = 1 - α, so that α, b are both positive numbers less than 1 with sum 1. Let g α (x) = sinh αx/sinh x, and f(x) = a g α (x) + b g β (x). For large positive x, g α (x) is approximately e αx /e x = e -βx which is a decreasing function of x, so we suspect that g α (x) has a maximum at x = 0. To check we look at small x, where we have g α (x) = α - α(1 - α 2 )/6 x 2 . So we conjecture that g α (x) has a maximum at x = 0, and similarly, of course, g β (x). This would imply that f(x) also had a maximum at x = 0 and indeed was strictly monotonic decreasing for x > 0 (and strictly monotonic increasing for x  0, αx  a, just use the result with a and b interchanged. Hence the | | in the answer. Comment. I do not like this solution (which is adapted from the official solution in the American Mathematical Journal). It gives no real clue as to what is going on. Better motivated or more enlightening solutions would be welcome. 52nd Putnam 1991 © John Scholes jscholes@kalva.demon.co.uk 28 Sep 1999