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

Let $\cos\alpha=\frac{x}{2},$ $\cos\beta=\frac{y}{2}$ and $\cos\gamma=\frac{z}{2}.$
Thus, $x$, $y$ and $z$ are positives, $x^2+y^2+z^2+xyz=4$ and we need to prove that:
$$x^3+y^3+z^3+x^3y^3z^3\geq4.$$
Indeed, let $$f(x,y,z,\lambda)=x^3+y^3+z^3+x^3y^3z^3-4+\lambda(x^2+y^2+z^2+xyz-4).$$
Thus, in the minimum point we need $$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=\frac{\partial f}{\partial z}=\frac{\partial f}{\partial\lambda}=0,$$
which gives $$3x^2+3x^2y^3z^3+\lambda(2x+yz)=3y^2+3y^2x^3z^3+\lambda(2y+xz)=3z^2+3z^2x^3y^3+\lambda(2z+xy)=0,$$ which gives
$$\frac{x^2+x^2y^3z^3}{2x+yz}=\frac{y^2+y^2x^3z^3}{2y+xz}=\frac{z^2+z^2x^3y^3}{2z+xy}$$ or
$$(x-y)(2x^2y^2z^3(x+y)-z(x^2+xy+y^2)-2xy)=0,$$
$$(x-z)(2x^2z^2y^3(x+z)-y(x^2+xz+z^2)-2xz)=0$$ and
$$(y-z)(2y^2z^2x^3(y+z)-x(y^2+yz+z^2)-2yz)=0.$$
Now, let $(x-y)(x-z)(y-z)\neq0.$
Thus, $$\frac{z(x^2+xy+y^2)+2xy}{z(x+y)}=\frac{y(x^2+xz+z^2)+2xz}{y(x+z)}=\frac{x(y^2+yz+z^2)+2yz}{x(y+z)},$$ which gives $$(x-y)(2z+xy)(xy+xz+yz)=0,$$
$$(x-z)(2y+xz)(xy+xz+yz)=0$$ and $$(y-z)(2x+yz)(xy+xz+yz)=0,$$ which is impossible.
Id est, $(x-y)(x-z)(y-z)=0$.
Let $y=x$.
Thus, the condition gives $$2x^2+z^2+x^2z=4$$ or
$$x^2(2+z)=4-z^2$$ or
$$z=2-x^2,$$ where $0<x<\sqrt2$ and we need to prove that
$$2x^3+(2-x^2)^3+x^6(2-x^2)^3\geq4$$ or
$$(x-1)^2(1+x-x^2)(x^8+3x^7+x^6-4x^5-4x^4-2x^3+4x+4)\geq0,$$ which is true because $$x^8+3x^7+x^6-4x^5-4x^4-2x^3+4x+4>0$$ even for any real $x$.
Also, we need to check, what happens for $xyz\rightarrow0^+$.
Let $z\rightarrow0^+$.
Thus, $x^2+y^2=4$ and
$$x^3+y^3\geq2\left(\sqrt{\frac{x^2+y^2}{2}}\right)^3=4\sqrt2>4$$ and we are done.
