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
$$\frac{1}{2}+64\cos^3{A}\cos^3{B}\cos^3{C}\ge 2\cos{A}\cos{B}\cos{C}+\dfrac{1}{4}(\cos{A}+\cos{B}+\cos{C})$$

The Euler's identity states that for all $x,y,z$ reals we have
$$x^3+y^3+z^3=3xyz+(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$
But if $A,B,C$ angles of an acute triangle $ABC$ and $x=\cos A$, $y=\cos B$, $z=\cos C$, we have
$$
x+y+z=1+\frac{\rho}{R},\tag 1
$$
$$
xyz=\frac{\rho_0}{2R}\tag 2
$$
$$
x^2+y^2+z^2=1-\frac{\rho_0}{R}\tag 3
$$
$$
xy+yz+zx=\frac{1}{2}\left(\frac{\rho}{R}\right)^2+\frac{\rho}{R}+\frac{1}{2}\frac{\rho_0}{R}\tag 4
$$
The symbols $R,\rho$ are the circumradius and inradius of $ABC$ resp. The symbol $\rho_0$ is the inradius of triangle $A'B'C'$. The $A'B'C'$ (orthic triangle) is formed by the intersections of heights of $ABC$ with its sides. Hence
$$
\Pi=\sum_{cyc}x^3+64(xyz)^3=
$$
$$
=3\frac{\rho_0}{2R}+\left(1+\frac{\rho}{R}\right)\left(1-\frac{\rho_0}{R}-\frac{1}{2}\left(\frac{\rho}{R}\right)^2-\frac{\rho}{R}-\frac{1}{2}\frac{\rho_0}{R}\right)+64\left(\frac{\rho_0}{2R}\right)^3=
$$
$$
=1-\frac{1}{2}\left(\frac{\rho}{R}\right)^3+8\left(\frac{\rho_0}{R}\right)^3-\frac{3}{2}\left(\frac{\rho}{R}\right)^2-\frac{3}{2}\frac{\rho}{R}\cdot\frac{\rho_0}{R}=
$$
$$
=f(t,t_0):=1-\frac{3t^2}{2}-\frac{t^3}{2}-\frac{3tt_0}{2}+8t_0^3\textrm{, }(t,t_0)\in D=\left[0,\frac{1}{2}\right]\times \left[0,\frac{1}{4}\right],
$$
where (Euler's theorem)
$$
t=\frac{\rho}{R}\leq\frac{1}{2}\textrm{ and }t_0=\frac{\rho_0}{R}\leq\frac{1}{4} 
$$
and $0<t\leq\frac{1}{2}$, $0<t_0\leq \frac{1}{4}$. One can see easily that
$$
f(t,t_0)\geq\frac{1}{2}=f\left(\frac{1}{2},\frac{1}{4}\right)\textrm{, }\forall (x,x_0)\in D
$$
This can be done easily and the proof is complete.
