If $A+B+C=\pi$, prove that $\cos (A-B) \cos (B-C) \cos (C-A)\ge 8\cos A \cos B \cos C$ If $A+B+C=\pi$, prove that $\cos (A-B) \cos (B-C) \cos (C-A)\ge 8\cos A \cos B \cos C$
I know this is true for acute angle triangle. 
I want to know whether it is true for every real $A,B,C$ such that $A+B+C=\pi.$
 A: Yes, this inequality is true for any reals $A,$ $B$ and $C$ such that $A+B+C=\pi.$
Indeed, let $\cos(A-B)=x$ and $\cos(A+B)=y.$
Thus, we need to prove that:
$$x(\cos(A+B-2C)+\cos(A-B))\geq8(\cos(A+B)+\cos(A-B))(-\cos(A+B))$$ or
$$x(\cos3(A+B)+x)+8y(x+y)\geq0$$ or
$$x(4y^3-3y+x)+8y(x+y)\geq0$$ or
$$x^2+(4y^3+5y)x+8y^2\geq0,$$ for which it's enough to prove that $f(x)\geq0,$ where
$$f(x)=x^2-(4y^3+5y)x+8y^2$$ and $\{x,y\}\subset[0,1].$
Now, if $$(4y^3+5y)^2-4\cdot8y^2\leq0$$ or
$$0\leq y\leq\frac{\sqrt{4\sqrt2-5}}{2},$$ so our inequality is proven.
But for $\frac{\sqrt{4\sqrt2-5}}{2}<y\leq1$ we see that $\frac{4y^3+5y}{2}>1,$ which says that $f$ decreases.
Id est, $$f(x)\geq f(1)=1-4y^3-5y+8y^2=(1-y)(2y-1)^2\geq0$$ and we are done!
A: Result to be established :
$$\begin{matrix}A+B+C=\pi \ \implies\\ \ \cos (A-B) \cos (B-C) \cos (C-A)\ge 8\cos A \cos B \cos C\end{matrix}\tag{*}$$
I would like to give here a variation on the excellent idea of Michael to use the following parameterization of a "triangle shape", i.e., a triangle known by its angles) :
$$x:=\cos(A-B), \ \ \ y=\cos(A+B)\tag{1}$$
Let me take his proof where
$$x^2+(4y^3+5y)x+8y^2 \ \ \text{has to be proven} \geq 0,\tag{2}$$
and take now a different path.
Result $(*)$ has been proven for acute angles ; we can consider WLOG (due to the exchangeability of $A,B,C$ in $(*)$ that 
$$\pi > A \geq \tfrac{\pi}{2} \geq B \geq C > 0\tag{3}$$
A first consequence of (3) is that 
$$C \leq \pi/2-A/2 \tag{&}$$
as shown by reasoning by contradiction. 
In order to understand the impact of restriction (3), I made a simulation (see figure below) that has evidenced that points $(x,y)$ defined by (1) are restricted to be in a certain narrow area bounded in particular by a curve (in red) whose non-evident parametrized (resp. cartesian) equation is (see explanation below)
$$\begin{cases}x&=&&\sin(3A/2)\\y&=&-&\sin(A/2)\end{cases} \ \ \implies \ \  x=4y^3-3y \tag{4} $$

As we have, for all $(x,y)$ :
$$-1 \leq x \leq 4y^3-3y\tag{5},$$
we can say that : $4y^3 \geq x+3y$, allowing to conclude, from (2):

$$x^2+(4y^3+5y)x+8y^2 \geq x^2+(x+3y+5y)x+8y^2=2(x+2y)^2 \geq 0$$

(with equality if and only if $x \to 1,y \to -1/2$ corresponding to the limit case where $A=B \to \pi/2$ whereas $C \to 0.$). 
Explanation for (4) :
For the limit case $C =(\pi-A)/2$ in (&), we have 


*

*$x=\cos(A-B)=\cos(A-(\pi-A-C))=\cos(2A+C-\pi)=$
$=-\cos (2A+C)=-\cos(\pi/2-3A/2)=\sin(3A/2)$ and 


*

*$y=\cos(A+B)=\cos(A+(\pi-A-C))=-\cos(C)=-\cos(\pi/2-A/2)=-\sin(A/2)$
