For $\triangle ABC$, prove $\sum_{cyc}\frac{1}{4+\cos A\cos(B-C)}\geq\frac23$ 
For the triangle $\triangle ABC$, prove the inquality,
  $$\frac1{4+\cos A\cos(B-C)}+\frac1{4+\cos B\cos(C-A)}+\frac1{4+\cos C\cos(A-B)}\ge \frac23$$
  where $A$, $B$ and $C$ are the vertex angles. 

I had trouble tackling it. Have treated it as a minimization problem, subject to the constraint $A+B+C=\pi$. In principle, it could be solved with the Lagrange multiplier. However, the required algebraic work seems to be prohibitive. Wonder whether there is a more elegant path to show it.
 A: First, we use the trig identity $ \cos A \cos B = \frac{1}{2} (\cos(A+B) + \cos(A-B)) $: 
$$\frac1{4 + \cos A \cos(B-C)}+\frac1{4  +\cos B \cos(C-A)} + \frac1{4 + \cos C \cos(A-B)} \ge \frac23$$
$$ \Leftrightarrow \frac1{4 + \frac{1}{2} (\cos(A+B-C) + \cos(A-B+C))} + \frac1{4 + \frac{1}{2} (\cos(B+C-A) + \cos(B-C+A))} + \frac1{4 + \frac{1}{2} (\cos(C+A-B) + \cos(C-A+B))} \ge \frac23 $$
Since $A + B + C = \pi$, we get:
$$ \frac1{4 + \frac{1}{2} (\cos(\pi-2C) + \cos(\pi-2B))} + \frac1{4 + \frac{1}{2} (\cos(\pi-2A) + \cos(\pi-2C))} + \frac1{4 + \frac{1}{2} (\cos(\pi-2B) + \cos(\pi-2A))} \ge \frac23 $$
Using the trig identity $ \cos(\pi-x) = - \cos x$:
$$ \frac1{4 - \frac{1}{2} (\cos(2C) + \cos(2B))} + \frac1{4 - \frac{1}{2} (\cos(2A) + \cos(2C))} + \frac1{4 - \frac{1}{2} (\cos(2B) + \cos(2A))} \ge \frac23 $$
Applying the inequality $ \frac1a + \frac1b + \frac1c \ge \frac9{a+b+c}$:
$$ \frac1{4 - \frac{1}{2} (\cos(2C) + \cos(2B))} + \frac1{4 - \frac{1}{2} (\cos(2A) + \cos(2C))} + \frac1{4 - \frac{1}{2} (\cos(2B) + \cos(2A))} $$
$$ \ge \frac9{12 - (\cos(2A) + \cos(2B) + \cos(2C))} $$
Finally, use the inequality $ \cos(2A) + \cos(2B) + \cos(2C) \ge - \frac{3}{2}$ (which you can find the proof in here: Prove that $\cos(2a) + \cos(2b) + \cos(2c) \geq -\frac{3}{2}$ for angles of a triangle):
$$ \frac{9}{12 - (\cos(2A) + \cos(2B) + \cos(2C))} \geq \frac{9}{12 - (- \frac{3}{2})}  = \frac23 $$
The inequality holds when 
$$A = B = C = \frac{\pi}{3} $$
or ABC is an equilateral triangle
A: Using $\cos(B+C)=\cos(\pi-A)=?$
and Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$
$$2(4+\cos A\cos(B-C))$$
$$=8-\cos(B+C)\cos(B-C)$$
$$=8-(\cos^2B-\sin^2C)$$
$$=9-(\cos^2B+\cos^2C)$$
Now using AM -HM inequity,  
$$\dfrac{\sum_{\text{cyc}}\dfrac2{9-(\cos^2 B+\cos^2C)}}3$$
$$\ge\dfrac3{\dfrac{27-2(\cos^2A+\cos^2B+\cos^2C)}2}$$
The equality will occur if $9-(\cos^2 B+\cos^2C)=9-(\cos^2C+\cos^2A)=9-(\cos^2A+\cos^2B)$
$\iff\cos^2A=\cos^2B=\cos^2C$ 
$\iff\sin^2A=\sin^2B=\sin^2C$ 
which will occur if $A=B=C$ 
using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $
