Competition problem — Algebra — Maths Olympiad Prep

Library / /4 of 28

Algebra Difficulty 7.4 National Olympiad, round 2 Prove it Hong Kong

Let $f(x)$ be a monic cubic polynomial with $f(0) = -64$ and all roots of $f(x)$ are nonnegative real numbers. What is the largest possible value of $f(-1)$ ? (A polynomial is monic if its leading coefficient is 1.)

Solution

The largest possible value of $f(-1)$ is $-125$ .

Let $f(x) = (x-a)(x-b)(x-c)$ where $a, b, c \ge 0$ and $abc = -f(0) = 64$ . Then we have
$\begin{aligned} f(-1) &= - (1+a)(1+b)(1+c) \\ &= -1 - (a+b+c) - (ab+bc+ca) - abc \\ &\le -1 - 3\sqrt[3]{abc} - 3\sqrt[3]{a^2b^2c^2} - abc \\ &= -125. \end{aligned}$
Equality holds when $a = b = c = 4$ , i.e. $f(x) = (x-4)^3$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.