Inequality for cosines Is the following inequality in a triangle known?
$$4(\cos A + \cos B + \cos C) \le 3 + \cos \left(\frac{B-C}{2}\right) + \cos \left(\frac{C-A}{2}\right) + \cos \left(\frac{A-B}{2}\right)$$
It looks correct to me but I would appreciate if someone confirm it. 

$\cos A+\cos B+\cos C=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}+1-2\sin^2\frac{C}{2}$
$=2\sin\frac{C}{2}\cos\frac{A-B}{2}+1-2\sin^2\frac{C}{2}\quad$  as $\quad A+B+C=\pi$ $\implies\cos\frac{A+B}{2}=\sin\frac{C}{2}$
$\leq 2\sin\frac{C}{2}+1-2\sin^2\frac{C}{2}\quad$  as $\quad\cos\frac{A-B}{2}\leq 1$
$=-\frac{1}{2}(2\sin\frac{C}{2}-1)^2+1+\frac{1}{2}$
The maximum value will come if $2\sin\frac{C}{2}=1\quad$ and if $\quad\cos\frac{A-B}{2}=1$
or if $\quad C=\frac{\pi}{3}\quad$  and $\quad A=B$. Then $A+B=\frac{2\pi}{3}\implies A=B=\frac{\pi}{3}=C$
The maximum value of $\cos A+\cos B+\cos C\quad$ thus $\quad\frac{3}{2}$
(i)Now, $4(\cos A+\cos B+\cos C)=2(\cos A+\cos B+\cos C)+2(\cos A+\cos B+\cos C)$
$\leq 2(\cos A+\cos B+\cos C)+2.\frac{3}{2}$
$=3+(\cos A+\cos B)+(\cos B+\cos C)+(\cos C+\cos A)$
Now,  $\cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}=2\sin\frac{C}{2}\cos\frac{A-B}{2}$
So, the problem reduces to $\sum 2\sin\frac{C}{2}\cos\frac{A-B}{2}≤ \sum \cos\frac{A-B}{2}$ 
(ii) Now, $4(\cos A+\cos B+\cos C)=(\cos A+\cos B+\cos C)+3(\cos A+\cos B+\cos C)$
$\leq (\cos A+\cos B+\cos C)+3.\frac{3}{2}$
$=3+\frac{1}{2}\sum (\cos A+\cos B + 1)$
$=3+\frac{1}{2}\sum (2\sin\frac{C}{2}\cos\frac{A-B}{2} + 1)$
So, the problem reduces to
$\sum( \sin\frac{C}{2}\cos\frac{A-B}{2} + \frac{1}{2}) ≤ \sum \cos\frac{A-B}{2} $
I'm trying to prove this.
