Problem 1397

First of all, $0 = (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$, i.e.,
$$ab + bc + ca = -\frac{1}{2},$$
and so $a$, $b$, $c$ are the roots of the cubic $x^3 - \frac{1}{2}x - abc = 0$, and, by hypothesis, these are real. Hence the product of the local extrema of this cubic is non-positive. But these extrema occur when $x = \pm\sqrt{6}$. Hence the requirement is that
$$\left(\frac{1}{6\sqrt{6}} - \frac{1}{2\sqrt{6}} - abc\right)\left(-\frac{1}{6\sqrt{6}} + \frac{1}{2\sqrt{6}} - abc\right) \le 0,$$
which simplifies to
$$\left(-\frac{1}{3\sqrt{6}} - abc\right) \left(\frac{1}{3\sqrt{6}} - abc\right) \le 0, \quad \text{ i.e. } \quad (abc)^2 \le \frac{1}{54},$$
the desired result. (More directly, of course, one can achieve the same result by quoting the criterion for the roots of a cubic in normal form to be real.) If the equality occurs, then $0$ is either a local max or a local min, in which case the cubic has a double root. Say, $a = b$, $c = -2a$, whence $6a^2 = 1$ and so equality happens iff two of $a$, $b$, $c$ are equal to $\pm\frac{1}{\sqrt{6}}$, and the third is $\mp\frac{2}{\sqrt{6}}$.

Another way is to use the following well-known fact, which is an easy consequence of problem 4: Suppose $a$, $b$, $c$ are the roots of the cubic $x^3 - px - q$. Then
$$(a-b)^2 (b-c)^2 (c-a)^2 = 4p^3 - 27q^2.$$
This implies $4p^3 - 27q^2 \ge 0$. With $2p = a^2 + b^2 + c^2 = 1$ the result follows.