Irish Mathematical Olympiad

Notice that $b \neq 0$ or $d \neq 0$ as $a^2 + ac + c^2 + 1 = (a + c/2)^2 + (3/4)c^2 + 1 > 0$. Suppose that $c = 0$. By assumption, we then have $a \neq 0$ and the second equation becomes $2ab + ad = 0$, hence $b = -d/2$. Substituting this into the first equation gives $d^2 = (4/3)(a^2 + 1)$. So $b^2 = (a^2 + 1)/3$ and
$$(3a^2 - b^2 + 1)(3c^2 - d^2 + 1) = \frac{1}{9}(8a^2 + 2)(-1 - 4a^2) < 0,$$

$$(2a + c)b = -(a + 2c)d.$$
Since $b \neq 0$ or $d \neq 0$, and $(2a + c)(a + 2c) = 2(a + c)^2 + ac > 0$, we deduce that $bd < 0$. Consider the complex numbers $z = a + ib$ and $w = c + id$. Then
$$z^2 + zw + w^2 = (a^2 - b^2 + ac - bd + c^2 - d^2) + i(2ab + ad + bc + 2cd) = -1.$$

So
$$(z^3 + z) - (w^3 + w) = (z - w)(z^2 + zw + w^2 + 1) = 0.$$
Now
$$\begin{aligned} z^3 + z &= a(a^2 - 3b^2 + 1) + ib(3a^2 - b^2 + 1), \\ w^3 + w &= c(c^2 - 3d^2 + 1) + id(3c^2 - d^2 + 1). \end{aligned}$$
In particular $d(3c^2 - d^2 + 1) = b(3a^2 - b^2 + 1)$. But $bd < 0$. So
$$(3a^2 - b^2 + 1)(3c^2 - d^2 + 1) \le 0.$$