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. 
 A: $\cos A+\cos B+\cos C=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}+1-2\sin^2\frac{C}{2}$
$=2\sin\frac{C}{2}\cos\frac{A-B}{2}+1-2\sin^2\frac{C}{2}\quad$  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.
A: Excuse me if what I'm going to tell is too imprecise but an approach that might work is the following: define $x_1=\cos\frac{A}{2},x_2=\sin\frac{A}{2}$ and similarly $x_3,x_4$ for $B$ and $x_5,x_6$ for $C$. Using the formulas for the sine and cosine of the sum of two angles the inequality we want to prove is (if I didn't mess anything up)
$$
4(x_1^2-x_2^2+x_3^2-x_4^2+x_5^2-x_6^2)\leq 3+x_1x_3+x_2x_4+x_1x_5+x_2x_6+x_3x_5+x_4x_6.
$$
Consider then the function $f:\mathbb{R}^6\rightarrow\mathbb{R}$ defined by
$$
f(x_1,\ldots,x_6)=4(x_1^2-x_2^2+x_3^2-x_4^2+x_5^2-x_6^2)-(x_1x_3+x_2x_4+x_1x_5+x_2x_6+x_3x_5+x_4x_6).
$$
We want to maximise it with the constraint given by $A+B+C=\pi$, which can be expressed in terms of the $x_i$, although not uniquely I'm afraid. Then we could make use of Lagrange multipliers.
Also $f$ is obviously a homogeneous degree 2 polynomial, so it defines a (projective) quadric. I wonder if this might be of help.
EDIT: The constraints $\cos(A+B+C)=-1$ and $\sin(A+B+C)=0$ in terms of the $x_i$ are
$$
(x_1^2-x_2^2)(x_2^2-x_3^2)(x_5^2-x_6^2)-4(x_1^2-x_2^2)x_3x_4x_5x_6-4x_1x_2(x_3^2-x_4^2)x_5x_6-4x_1x_2x_3x_4(x_5^2-x_6^2)=-1
$$
and
$$
2x_1x_2(x_3^2-x_4^2)(x_5^2-x_6^2)+2(x_1^2-x_2^2)x_3x_4(x_5^2-x_6^2)+2(x_1^2-x_2^2)(x_3^2-x_4^2)x_5x_6-8x_1x_2x_3x_4x_5x_6=0.
$$
Note that this only tells us that $A+B+C$ is an odd multiple of $\pi$ and not exactly $\pi$.
A: 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 $A,B,C$.
A: Let $a=y+z$, $b=x+z$ and $c=x+y$.
Hence, we need to prove that
$$2\sum_{cyc}\frac{a^2+b^2-c^2}{ab}\leq3+\sum_{cyc}\left(\sqrt{\frac{(a+b+c)^2(b+c-a)(a+c-b)}{16c^2ab}}+\sqrt{\frac{(a+b-c)^2(b+c-a)(a+c-b)}{16c^2ab}}\right)$$ or
$$2\sum_{cyc}c(a^2+b^2-c^2)\leq3abc+\sum_{cyc}\frac{a+b}{2}\sqrt{ab(a+c-b)(b+c-a)}$$ or
$$4\sum_{cyc}(x+y)(z^2+xz+yz-xy)\leq$$
$$\leq3(x+y)(x+z)(y+z)+\sum_{cyc}(x+y+2z)\sqrt{xy(x+z)(y+z)}$$ or
$$\sum_{cyc}(x+y+2z)\sqrt{xy(x+z)(y+z)}\geq\sum_{cyc}(x^2y+x^2z+6xyz),$$
which is C-S and AM-GM:
$$\sum_{cyc}(x+y+2z)\sqrt{xy(x+z)(y+z)}\geq\sum_{cyc}(x+y+2z)\sqrt{xy}(\sqrt{xy}+z)=$$
$$=\sum_{cyc}(x^2y+x^2z+2xyz)+\sum_{cyc}(x+y+2z)z\sqrt{xy}\geq$$
$$=\sum_{cyc}(x^2y+x^2z+2xyz)+\sum_{cyc}4xyz=$$
$$=\sum_{cyc}(x^2y+x^2z+6xyz).$$
Done!
