Find min of $M=\frac{1}{2+\cos2A}+\frac{1}{2+\cos2B}+\frac{1}{2-\cos2C}$ 
Find min of $$M=\frac{1}{2+\cos2A}+\frac{1}{2+\cos2B}+\frac{1}{2-\cos2C}$$, where $A, B, C$ are three angle of triangle $ABC$


Using Cauchy-Schwarz, we obtain: \begin{align*}
M &= \frac{1}{2+\cos2A}+\frac{1}{2+\cos2B}+\frac{1}{2-\cos2C}\\ 
 &\ge \frac{9}{6+2\cos(A+B)\cos(A-B)+1-2\cos^2C}\\ 
 & \ge \frac{9}{7-2\cos C-2\cos^2C}\ge \frac{9}{\frac{15}{2}}=\frac{6}{5}
\end{align*}
The equality holds if and only if $\left\{\begin{matrix}A=B\\\cos C=-\frac{1}{2} \end{matrix}\right.\iff \left\{\begin{matrix}A=B=30^{\circ}\\ C=120^{\circ}\end{matrix}\right.$

But the change $\cos(A+B)\cos(A-B)\le-\cos C $ may be not correct. :(


The change might not be correct, but before that the numerator is 
$$
f(A,B)=6+\cos 2A+\cos 2B-\cos 2C=6+\cos 2A+\cos 2B-\cos 2(A+B)
$$
$$
\frac{\partial f}{\partial A}=2 \sin (2 A+2 B)-2 \sin (2 A)
$$
$$
\frac{\partial f}{\partial B}=2 \sin (2 A+2 B)-2 \sin (2 B)
$$
We can only have extrema at the border of the set, namely $A=0$, $B=0$, $A+B=\pi$ or where both partials are $0$, meaning $\sin 2A=\sin 2B$, equivalently $A=B$ or $A+B=k\frac{\pi}{2}$.
From checking these we get that the minimum is at $\frac{\pi}{6},\frac{\pi}{6}$ and is in fact $\frac{6}{5}$
