Putnam 1998, Problem A-3

Let \( f \) be a real function on the real line with continuous third derivative. Prove that there exists a point \( a \) such that
\[
f(a) \cdot f'(a) \cdot f''(a) \cdot f'''(a) \geq 0.
\]

Solution:
If at least one of \( f(a), f'(a), f''(a) \), or \( f'''(a) \) vanishes at some point \( a \), then we are done. Hence we may assume each of \( f(x), f'(x), f''(x) \), and \( f'''(x) \) is either strictly positive or strictly negative on the real line. By replacing \( f(x) \) by \( -f(x) \) if necessary, we may assume \( f''(x) > 0 \); by replacing \( f(x) \) by \( f(-x) \) if necessary, we may assume \( f'''(x) > 0 \). (Notice that these substitutions do not change the sign of \( f(x)f'(x)f''(x)f'''(x) \).) Now \( f''(x) > 0 \) implies that \( f'(x) \) is increasing, and \( f'''(x) > 0 \) implies that \( f'(x) \) is convex, so that \( f'(x + a) > f'(x) + a f''(x) \) for all \( x \) and \( a \). By letting \( a \) increase in the latter inequality, we see that \( f'(x + a) \) must be positive for sufficiently large \( a \); it follows that \( f'(x) > 0 \) for all \( x \). Similarly, \( f'(x) > 0 \) and \( f''(x) > 0 \) imply that \( f(x) > 0 \) for all \( x \). Therefore \( f(x)f'(x)f''(x)f'''(x) > 0 \) for all \( x \), and we are done.