Prove this inequality $\sum \cos{A}\ge\frac{1}{4}(3+\sum\cos{(A-B)})$ Prove that in any triangle $ABC$ the following inequality holds
$$\cos{A}+\cos{B}+\cos{C}\ge\dfrac{1}{4}(3+\cos{(A-B)}+\cos{(B-C)}+\cos{(C-A)})$$
And I have gotten
$$8(\cos{A}+\cos{B}+\cos{C})\ge 6+2(\cos{(A-B)}+\cos{(B-C)}+\cos{(C-A)})$$
$$2(\cos{(A-B)}+\cos{(B-C)}+\cos{(C-A)})+3=(\sum_{cyc}\cos{A})^2+(\sum_{cyc}\sin{A})^2$$
$$\Longleftrightarrow 8\sum_{cyc}\cos{A}\ge 3+(\sum_{cyc}\cos{A})^2+(\sum_{cyc}\sin{A})^2$$
then Any hints, ideas? Thanks in advance.

use 
$$\sum\cos{A}=\dfrac{R+r}{R},\sum\cos{A}\cos{B}=\dfrac{s^2+r^2-4R^2}{4R^2},\sum\sin{A}\sin{B}=\dfrac{s^2+4Rr+r^2}{4R^2}$$
$$\Longleftrightarrow \dfrac{4R+4r}{R}\ge 3+\dfrac{s^2+r^2-4R^2}{4R^2}+\dfrac{s^2+4Rr+r^2}{4R^2}$$
$$\Longleftrightarrow 4R^2+6Rr\ge s^2+r^2$$
use Gerrentsen inequality
$$s^2\le 4R^2+4Rr+3r^2$$
we only prove following
$$4R^2+6Rr\ge 4R^2+4Rr+4r^2$$
it equal to Euler inequality
$$R\ge 2r$$
