Is the smallest root $x$ of
$$ 10x^{3}-30x^{2}+\left(30-2\sum_{1\le i<j\le6}\cos^{2}\alpha_{ij}\right)x\\ +2\sum_{1\le i<j\le6}\cos^{2}\alpha_{ij}-\sum_{1\le i<j<k\le6}\cos\alpha_{ij}\cos\alpha_{ik}\cos\alpha_{jk}-10 $$ never larger than $\frac{1}{2}$ for all $\alpha_{ij}$ being the angles between 6 rays in $\mathbb{R}^{3}$?
Any helpful answer would be highly appreciated.
$\newcommand\al\alpha$Let $$u:=\sum_{1\le i<j\le6}c_{ij}^2\quad\text{and}\quad v:=\sum_{1\le i<j<k\le6}c_{ij}c_{ik}c_{jk},$$ where $c_{ij}:=\cos\al_{ij}$. This answer is somewhat similar to the previous answer -- to a much simpler question.
Let $p(x)$ be the polynomial in question. Note that $0\le u\le\binom62=15$ and let $$x_*:=1-\sqrt{u/15}.$$ According to this answer by Fedor Petrov, $$v\le\frac4{3\sqrt{15}}\,u^{3/2}.$$ So, $$p(x_*)=\frac4{3\sqrt{15}}\,u^{3/2}-v\ge0.$$ Also, $p(-\infty+)=-\infty<0$. So, $p$ has a root $\le1/2$ in the case when $x_*\le1/2$.
It remains to consider the case when $x_*>1/2$, that is, when $0\le u<15/4$. Note that $$p(1/2)=-5/4+u-v.$$ According to this answer by GH from MO, $u-v\ge23/8$ if $u\le15/4$. So, in the case when $x_*>1/2$ we have $$p(1/2)\ge-5/4+23/8=13/8\ge0.$$ So, $p$ has a root $\le1/2$ in the case when $x_*>1/2$ as well.
We also see that it was enough to prove that $u-v\ge10/8$ (given $u\le15/4$), which seems a much weaker claim than $u-v\ge23/8$ (given $u\le15/4$).
I will show more, namely that the 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 the nonnegative and nonpositive parts of the three coordinate axes, 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 question is $$p(x):=10 x^3 - 30 x^2 + (30 - 2 u) x + (2 u - v - 10).$$ The discriminant of this polynomial equals $20(16 u^3 - 135 v^2)$, hence it is nonnegative by this answer of Fedor Petrov. 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, $$(u-3)^3\geq 27(u-v-3)^2>27\left(2\sqrt{\frac{3}{5}}-1\right)^2(u-3)^2.$$ Comparing the two sides, we obtain a contradiction: $$6>u-3>27\left(2\sqrt{\frac{3}{5}}-1\right)^2>8.$$
