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
$$\frac{1}{2}+64\cos^3{A}\cos^3{B}\cos^3{C}\ge 2\cos{A}\cos{B}\cos{C}+\dfrac{1}{4}(\cos{A}+\cos{B}+\cos{C})$$
 A: 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.
A: Problem: Let $a, b, c > 0$ with $a^2+b^2+c^2+abc=4$. Prove that $a^3+b^3+c^3+(abc)^3\ge 4$.
Solution: It suffices to prove that, for $a, b, c > 0$,
$$a^3+b^3+c^3+(abc)^3 - 4 - 2(a^2+b^2+c^2+abc - 4) \ge 0. \tag{1}$$
It is verified by Mathematica.
By Vasc's Equal Variable Theorem [1, Corollary 1.9], we only need to prove the case when
$a=b$. The proof is ugly. Omitted. If I find nice proofs, I will update the post.
I hope to see simple SOS (Sum of Squares) solutions of (1).
Reference
[1] Vasile Cirtoaje, “The Equal Variable Method”, J. Inequal. Pure and Appl. Math., 8(1), 2007.
% https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf
Remark: The stronger version is also true: for $a, b, c > 0$,
$$a^3+b^3+c^3+2(abc)^2 - abc - 4 - 2(a^2+b^2+c^2+abc - 4) \ge 0. \tag{2}$$
(Note: $(abc)^3 + abc \ge 2(abc)^2$.)
A: 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.
A: Another way.
Let $a^2+b^2-c^2=z$, $a^2+c^2-b^2=y$ and $b^2+c^2-a^2=x$.
Thus, $x$, $y$ and $z$ are positives, $\cos\alpha=\frac{x}{\sqrt{(x+y)(x+z)}},$ $\cos\beta=\frac{y}{\sqrt{(x+y)(y+z)}}$, $\cos\gamma=\frac{z}{\sqrt{(x+z)(y+z)}}$
and we need to prove that
$$2\sum_{cyc}\frac{x^3}{\sqrt{(x+y)^3(x+z)^3}}+\frac{128x^3y^3z^3}{\prod\limits_{cyc}(x+y)^3}\geq1.$$
Now, by AM-GM  $$\sum_{cyc}\frac{x^3}{\sqrt{(x+y)^3(x+z)^3}}=\sum_{cyc}\frac{2x^3}{2(x+y)(x+z)\sqrt{(x+y)(x+z)}}\geq$$
$$\geq \sum_{cyc}\frac{2x^3}{(x+y)(x+z)(2x+y+z)}.$$
Id est, it's enough to prove that:
$$ \sum_{cyc}\frac{4x^3}{(x+y)(x+z)(2x+y+z)}+\frac{128x^3y^3z^3}{\prod\limits_{cyc}(x+y)^3}\geq1,$$ which is obvious by BW (https://artofproblemsolving.com/community/c6h522084 ).
By the way, a full expanding gives $$\sum_{sym}(2x^9y^3+7x^8y^4+10x^7y^5+5x^6y^6)+$$
$$+xyz\sum_{sym}(6x^8y+28x^7y^2+30x^6y^3-8x^5y^4)+$$
$$+x^2y^2z^2\sum_{sym}(21x^6+40x^5y-154x^4y^2-158x^3y^3)+$$
$$+x^3y^3z^3\sum_{sym}(-8x^3+58x^2y+121xyz)\geq0,$$
which is not so trivial.
A: 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}
