I will show more, namely that the cubic polynomial in question has three real roots, the smallest of which is at most
$$d:=\frac{5-\sqrt{15}}{5}=0.22540333\dotsc$$
This estimate is sharp, because in case the underlying six rays are $$(\pm 1,0,0)\mathbb{R}_{\geq 0},\qquad (0,\pm 1,0)\mathbb{R}_{\geq 0},\qquad (0,0,\pm 1)\mathbb{R}_{\geq 0},$$
three of the angles $\alpha_{ij}$ ($i<j$) are equal to $\pi$, twelve of the angles $\alpha_{ij}$ ($i<j$) are equal to $\pi/2$, and the polynomial is
$$10x^3-30x^2+24x-4=20\left(x-1\right)\left(x-\frac{5-\sqrt{15}}{5}\right)\left(x-\frac{5+\sqrt{15}}{5}\right).$$
We turn to the proof. With the notation
$$u:=\sum_{1\leq i<j\leq 6}\cos^{2}\alpha_{ij}\qquad\text{and}\qquad
v:=\sum_{1\leq i<j<k\leq 6}\cos\alpha_{ij}\cos\alpha_{ik}\cos\alpha_{jk},$$
the polynomial in the original post equals
$$p(x):=10 x^3 - 30 x^2 + (30 - 2 u) x + (2 u - v - 10)$$
The discriminant of this polynomial, $20(16 u^3 - 135 v^2)$, is nonnegative by [this answer of Fedor Petrov][1]. So $p(x)$ has three real roots (counted with multiplicity). Let us assume that all these roots exceed $d$. Then the three roots of
$$p(d+x)=10x^3-6\sqrt{15}x^2+2\left(9-u\right)x+\left(2\sqrt{\frac{3}{5}}u-v-6\sqrt{\frac{3}{5}}\right)$$
are positive, hence in particular
$$u<9\qquad\text{and}\qquad 2\sqrt{\frac{3}{5}}(u-3)<v.$$
Using also $(\ast)$ from my response [for this other MO question][2],
$$(u-3)^3\geq 27(u-v-3)^2>27\left(2\sqrt{\frac{3}{5}}-1\right)^2(u-3)^2.$$
Comparing the two sides, we obtain a contradiction:
$$6>u-3>27\left(2\sqrt{\frac{3}{5}}-1\right)^2>8.$$

  [1]: https://mathoverflow.net/a/479088/36721
  [2]: https://mathoverflow.net/q/479107/