R+ is the positive reals, f : [0, 1] → R+ is monotonic decreasing. Show that ∫01 f(x) dx ∫01 x f(x)2 dx ≤ ∫01 x f(x) dx ∫01 f(x)2 dx.
Solution
(f(x) - f(y) )(y - x) ≥ 0. Also f(x), f(y) ≥ 0, so ∫ ∫ f(x) f(y) (f(x) - f(y) )(y - x) dx dy ≥ 0. But f(x) f(y) (f(x) - f(y) )(y - x) = ( f(x)2 y f(y) - f(x) y f(y)2 ) + ( f(y)2 x f(x) - f(y) x f(x)2). So if we take the same limits of integration, 0 and 1, for both integrals, then we have ∫ ∫ f(x) f(y) (f(x) - f(y) )(y - x) dx dy = 2 ∫ f(x)2 y f(y) - f(x) y f(y)2 dx dy = 2 ∫ f(x)2 dx ∫ y f(y) dy - 2 ∫ f(x) dx ∫ y f(y)2 dy.
© John Scholes
jscholes@kalva.demon.co.uk
5 Mar 2002