Irska

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 $$\left(-\frac{1}{3\sqrt{6}}-abc\right)\left(\frac{1}{3\sqrt{6}}-abc\right)\le 0,\text{ i.e. }(abc{)}^2\le \frac{1}{54},$$ 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.