Irish Mathematical Olympiad

Suppose $(m, c)$ is an allowable pair. Since the inequality must hold when $x = 0$, we see immediately that $c$ cannot be positive. Moreover,
$$m \le \frac{x^3 - c}{x}, \quad \forall x > 0,$$
and so, if $c = 0$, then $m \le 0$. Otherwise, $c < 0$, and so, by the AM-GM inequality,
$$\begin{aligned} \frac{x^3 - c}{x} &= x^2 + \frac{(-c)}{x} = x^2 + \frac{(-c)}{2x} + \frac{(-c)}{2x} \\ &\ge 3\sqrt[3]{x^2 \frac{(-c)}{2x} \frac{(-c)}{2x}} = 3\sqrt[3]{\frac{c^2}{4}}, \end{aligned}$$
with equality iff
$$x^2 = \frac{-c}{2x}, \text{ i.e. } c = -2x^3.$$
Hence, using this value of $x$ we obtain,
$$m \le \frac{x^3 - c}{x} = 3x^2$$
Thus, we have shown that either
(i) $c = 0$ and $m \le 0$, or
(ii) $c = -2x^3$ for some $x > 0$, and $m \le 3x^2$.

Conversely, suppose $m \le 3s^2$, $c = -2s^3$ for some $s \ge 0$. Then for all $x \ge 0$
$$mx + c \le 3s^2x - 2s^3.$$ But $3s^2x - 2s^3 \le x^3$ is equivalent to
$$0 \le x^3 - 3s^2x + 2s^3 = (x - s)(x^2 + sx - 2s^2) = (x - s)^2(x + 2s),$$
which is true for all $x \ge 0$. It follows that
$$\begin{aligned} & \{(m, c) : c = -2s^3, m \le 3s^2, \text{ for some } s \ge 0\} \\ &= \{(m, c) : c \le 0, 27c^2 - 4m^3 \ge 0\} \end{aligned}$$
is the set of allowable pairs.