In acute angle triangle, prove that $\cos (A-B) \cos (B-C) \cos (C-A)\ge 8\cos A \cos B \cos C$

Question

In acute angle triangle, prove that $$\cos (A-B) \cos (B-C) \cos (C-A)\ge 8\cos A \cos B \cos C$$

My progress:

$$\cos(A-B)=2\cos A\cos B+\cos C\ge 2\sqrt{2\cos A\cos B\cos C}$$ $$\implies \prod\cos(A-B)\ge 8\left (2\prod \cos A\right )^{\frac 32}$$

Answer 1

$$2\sin A\cos(B-C)=\sin2B+\sin2C\ge2\sqrt{\sin2B\sin2C}=4\sqrt{\sin B\cos B\sin C\cos C}$$

As $A,B,C$ are acute, all the trigonometric ratios are $>0$ $$\implies\cos(B-C)\ge?$$

Similarly for $\cos(C-A),\cos(A-B)$