prove $\sum\cos^3{A}+64\prod\cos^3{A}\ge\frac{1}{2}$ In every acute-angled triangle $ABC$,show that
$$(\cos{A})^3+(\cos{B})^3+(\cos{C})^3+64(\cos{A})^3(\cos{B})^3(\cos{C})^3\ge\dfrac{1}{2}$$
I want use Schur inequality
$$x^3+y^3+z^3+3xyz\ge xy(y+z)+yz(y+z)+zx(z+x)$$
then we have
$$x^3+y^3+z^3+6xyz\ge (x+y+z)(xy+yz+zx)$$
But I can't use this to prove my question
and I use this post methods  links also can't solve my problem,use $AM-GM $ inequality$$\cos^3{A}+\dfrac{\cos{A}}{4}\ge\cos^2{A}$$
so
$$LHS\ge \sum_{cyc}\cos^2{A}-\dfrac{1}{4}\sum_{cyc}\cos{A}+64\prod_{cyc}\cos^3{A}$$use
$$\cos^2{A}+\cos^2{B}+\cos^2{C}+2\cos{A}\cos{B}\cos{C}=1$$
it must to prove
$$\frac{1}{2}+64\cos^3{A}\cos^3{B}\cos^3{C}\ge 2\cos{A}\cos{B}\cos{C}+\dfrac{1}{4}(\cos{A}+\cos{B}+\cos{C})$$

HINT
Firstly, let
$$\tan \dfrac A2 = x,\quad \tan \dfrac B2 = y,\quad s=(x+y)^2,\quad p=xy,\quad x,y \in(0,1),\tag1$$
then
\begin{align}
&\tan\dfrac C2 = \cot\left(\dfrac A2+\dfrac B2\right) = \dfrac{1-xy}{x+y} = \dfrac{1-p}{\sqrt s}\in(0,1),\\[4pt]
&\cos A = \dfrac{1-x^2}{1+x^2},\quad \cos B = \dfrac{1-y^2}{1+y^2},\quad 
\cos C = \dfrac{s-(1-p)^2}{s+(1-p)^2} > 0,\\[4pt]
&\cos A + \cos B =  \dfrac{(1-x^2)(1+y^2)+(1+x^2)(1-y^2)}{(1+x^2)(1+y^2)}
= \dfrac{2-2p^2}{s+(1-p)^2},\\[4pt]
&\cos A\cos B =  \dfrac{(1-x^2)(1-y^2)}{(1+x^2)(1+y^2)}
= \dfrac{(1+p)^2-s}{s+(1-p)^2},\\[4pt]
&s \ge 2p,\quad \sqrt s>1-p,\quad s+2\sqrt s>2,
\end{align}
\begin{cases}
s\in(0,4)\\[4pt]
p\in\left(|1-\sqrt s|,\dfrac s4\right].\tag2
\end{cases}
Secondly,
\begin{align}
&\cos A + \cos B + \cos C = \dfrac{s+(1-p)(2+2p-1+p)}{s+(1-p)^2}
 = 1+\dfrac{4p(1-p)}{s+(1-p)^2},\\[4pt]
&\cos A\cos B\cos C = \dfrac{((1+p)^2-s)(s-(1-p)^2)}{(s+(1-p)^2)^2},\\[4pt]
&\cos A + \cos B + \cos C + 4\cos A \cos B \cos C \\[4pt]
&= 1+\dfrac{4\big(p(1-p)(s+(1-p)^2)+((1+p)^2-s)(s-(1-p)^2)\big)}{(s+(1-p)^2)^2} \ge 2 (?)\\[4pt]
\end{align}
Then should
\begin{align}
&F = \cos^3A+\cos^3B+\cos^3C + 64\cos^3A\cos^3B\cos^3C\\[4pt]
&\ge \dfrac1{16}(\cos A+\cos B+\cos C + 4\cos A\cos B\cos C)^3\ge \dfrac12.
\end{align}
