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. 

WLOG let $C$ be the largest angle, let
$x=\frac{A-B}{2}, y=\frac{A+B}{2}$ then $-\frac{\pi}{4}<x<\frac{\pi}{4}, 0<y<\frac{\pi}{2}$. 
$$
A = x+y,~~ B=y-x,~~ C = \pi-2y \\
\cos\left(\frac{B-C}{2}\right) = \sin\left(\frac{3y-x}{2}\right), ~~
\cos\left(\frac{C-A}{2}\right) = \sin\left(\frac{3y+x}{2}\right) \\
\sin\left(\frac{3y-x}{2}\right)+\sin\left(\frac{3y+x}{2}\right) = 2\cos(x/2)\sin(3y/2) \\
\cos(x/2)\ge \cos x>\frac{1}{\sqrt{2}}, ~\sin(3y/2)>0 \Rightarrow 2\cos(x/2)\sin(3y/2)\ge2\cos x\sin(3y/2)
$$
Then we can write
$$
\begin{align}
3 \!&+\cos\left(\frac{A-B}{2}\right)+\cos\left(\frac{B-C}{2}\right)+\cos\left(\frac{C-A}{2}\right) -4(\cos A + \cos B + \cos C) \\
& = 3 + \cos x + 2\cos(x/2)\sin(3y/2)-4(\cos(x+y)+\cos(y-x)-\cos2y) \\
& = -1+\cos x+2\cos(x/2)\sin(3y/2)-8\cos x\cos y+8\cos^2y \\
& \ge 8\cos^2y-1+\cos x(1+2\sin(3y/2)-8\cos y) \\
& = 8\cos^2y-1+\cos x \cdot f(y)
\end{align}
$$
where we define $f(y)=1+2\sin(3y/2)-8\cos y$.
Writing $v=y/2$ we can work through
$$
\begin{align}
8\cos^2y-1+f(y) & = 7-8\sin^22v+1+2\sin3v-8\cos 2v\\
& = 8-16\sin^2v(1-2\sin^2v)+6\sin v-8\sin^3v-8(1-2\sin^2v) \\
& = 32\sin^4v-8\sin^3v-16\sin^2v+6\sin v \\
& = 2\sin v(1-2\sin v)^2(3+4\sin v)
\end{align}
$$
This last expression is clearly $\ge 0$ when $\sin v\ge 0$, but to answer the original question we now need to consider cases.
If $f(y)\le 0$, which corresponds to $0<y\le R=1.192797\cdots$ then
$$
8\cos^2y-1+\cos x\cdot f(y)\ge 8\cos^2y-1+f(y) \ge 0
$$
as we just described.
On the other hand if $R<y<\pi/2, ~f(y)>0$ then since $|x|<\pi/4$
$$
\begin{align}
8\cos^2y-1+\cos x\cdot f(y) & \ge 8\cos^2y-1+\frac{1}{\sqrt{2}}f(y)\\
& = 8\cos^2y-1+f(y)-f(y)\left(1-\frac{1}{\sqrt{2}}\right) \\
& = 2\sin v(1-2\sin v)^2(3+4\sin v)-f(y)\left(1-\frac{1}{\sqrt{2}}\right)
\end{align}
$$
I don't have an algebraic way to proceed from here, but it is easy to check numerically that the last expression is positive for $y$ in the desired range, which establishes the original inequality in $A,B,C$.
