Return to Answer

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 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 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 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: $$6>u-3>27\left(2\sqrt{\frac{3}{5}}-1\right)^2>8.$$

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: $$6>u-3>27\left(2\sqrt{\frac{3}{5}}-1\right)^2>8.$$

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. 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: $$6>u-3>27\left(2\sqrt{\frac{3}{5}}-1\right)^2>8.$$

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 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 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 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: $$6>u-3>27\left(2\sqrt{\frac{3}{5}}-1\right)^2>8.$$

I will show a bit more, namely that the cubic polynomial in question has three real roots, the smallest of which is at most $$u:=\frac{5-\sqrt{15}}{5}=0.22540333\dotsc$$

Let $u$ and $v$ be as in Iosif Pinelis's response. Then the cubic polynomial in question equals $$p(x):=10 x^3 - 30 x^2 + (30 - 2 u) x + (2 u - v - 10),$$ and its discriminant $20(16 u^3 - 135 v^2)$ 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: $$6>u-3>27\left(2\sqrt{\frac{3}{5}}-1\right)^2>8.$$

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. 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: $$6>u-3>27\left(2\sqrt{\frac{3}{5}}-1\right)^2>8.$$