3rd Putnam 1940 Problem B7 Given n > 8, let a = √n and b = √(n+1). Which is greater a b or b a ? Solution Straightforward
: a b is greater. a b = e b ln a and b a = e a ln b . So we have to decide which of b ln a and a ln b is greater, or, equivalently, which of (ln a)/a and (ln b)/b is greater. The latter is clearly more promising. So set f(x) = (ln x)/x. Then f '(x) = 1/x 2 - (ln x)/x 2 which is negative for x > e. Obviously b > a, so provided a > e, (ln a)/a > (ln b)/b and hence b ln a > a ln b and a b > b a . But e 2 < 9, so the result is certainly true for n ≥ 9. 3rd Putnam 1940 © John Scholes jscholes@kalva.demon.co.uk 15 Sep 1999