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Iosif Pinelis
Iosif Pinelis
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$\newcommand\al\alpha$This answer is incomplete -- it reduces the problem to a conjecture involving, not $x$, but onlyLet $$u:=\sum_{1\le i<j\le6}c_{ij}^2\quad\text{and}\quad v:=\sum_{1\le i<j<k\le6}c_{ij}c_{ik}c_{jk},$$ where $c_{ij}:=\cos\al_{ij}$. This answer is somewhat similar to the previous answer -- to a much simpler question (the previous answer was complete).

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 conjecturethis 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, provided the mentioned conjecture is true, $p$ will havehas a root $\le1/2$ in the case when $x_*>1/2$ as well.

We also see that it would bewas enough to prove that $u-v\ge10/8$ (given $u\le15/4$), which seems a much weaker claim than $u-v\ge23/8$ (given $u\le15/4$).

$\newcommand\al\alpha$This answer is incomplete -- it reduces the problem to a conjecture involving, not $x$, but only $$u:=\sum_{1\le i<j\le6}c_{ij}^2\quad\text{and}\quad v:=\sum_{1\le i<j<k\le6}c_{ij}c_{ik}c_{jk},$$ where $c_{ij}:=\cos\al_{ij}$. This answer is somewhat similar to the previous answer -- to a much simpler question (the previous answer was complete).

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 conjecture, $u-v\ge23/8$ 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, provided the mentioned conjecture is true, $p$ will have a root $\le1/2$ in the case when $x_*>1/2$ as well.

We also see that it would be enough to prove that $u-v\ge10/8$ (given $u\le15/4$), which seems a much weaker claim than $u-v\ge23/8$ (given $u\le15/4$).

$\newcommand\al\alpha$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 $u\le15/4$).

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Iosif Pinelis
Iosif Pinelis
  • 127.8k
  • 8
  • 107
  • 229

$\newcommand\al\alpha$This answer is incomplete -- it reduces the problem to a conjecture involving, not $x$, but only $$u:=\sum_{1\le i<j\le6}c_{ij}^2\quad\text{and}\quad v:=\sum_{1\le i<j<k\le6}c_{ij}c_{ik}c_{jk},$$ where $c_{ij}:=\cos\al_{ij}$. This answer is somewhat similar to the previous answer -- to a much simpler question (the previous answer was complete).

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 conjecture, $u-v\ge23/8$ 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, provided the mentioned conjecture is true, $p$ will have a root $\le1/2$ in the case when $x_*>1/2$ as well.

We also see that it would be enough to prove that $u-v\ge10/8$ (given $u\le15/4$), which seems a much weaker claim than $u-v\ge23/8$ (given $u\le15/4$).