### 60th Putnam 1999

Problem B4

Let R be the reals. f :R → R is three times differentiable, and f(x), f '(x), f ''(x), f '''(x) are all positive for all x. Also f(x) ≥ f '''(x) for all x. Show that f '(x) < 2 f(x) for all x.

Solution

Take an arbitrary point x = a. Let f(a) = b, f '(a) = c, f ''(a) = d. We wish to show that c < 2b.

f ''(x) has positive derivative, so f ''(x) ≤ f ''(a) = d for x <= a. Hence f '(x) ≥ f '(a) + d(x - a) = c + d(x - a), and f(x) ≤ f(a) + c(x - a) + d(x - a)2/2 = b + c(x - a) + d(x - a)2/2 for x ≤ a. But f(x) is positive for x ≤ a, so the quadratic b + c(x - a) + d(x - a)2/2 has no roots for (x - a) negative. But it has no roots for positive (x - a) since all its coefficients are positive. Hence it has no real roots and so c2 < 2bd.

We are given that f '''(x) ≤ f(x), so f '''(x) < f(x) < f(a) = b for x < a. Hence f ''(x) ≥ f ''(a) + b(x - a) = d + b(x - a) for x ≤ a. Hence f '(x) ≤ f '(a) + d(x - a) + b(x - a)2/2 = c + d(x - a) + b(x - a)2/2 for x <= a. But f '(x) is everywhere positive, so the quadratic c + d(x - a) + b(x - a)2/2 has no roots for negative (x - a). It also has none for positive (x - a), so it has no real roots and d2 < 2bc.

So c4 < 4b2d2 < 8b3c, and hence c < 2b, which is the required result.