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 $A+B+C=\pi.$

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!
