Mongolian Mathematical Olympiad

First recall that $2\prod \cos \alpha =1-\sum {\cos }^2\alpha$. Now using the identity $\cos (3x)=4{\cos }^3(x)-3\cos (x)$, we see that $$\begin{array}{rl}4\sum \cos (\alpha )\cos (3\alpha )-8\prod \cos (\alpha )+7& =16\sum {\cos }^4(\alpha )-8\sum {\cos }^2(\alpha )+3\\ & =\sum (4{\cos }^2(\alpha )-1{)}^2\ge 0.\end{array}$$ The condition for equality is $\alpha =\beta =\gamma =\pi /3$.