Bulgarian Mathematical Olympiad

First solution. The inequality follows by the fact that if $\theta \in [0,\frac{\pi }{2}]$, then $$\cos 2\theta =1-\sin 2\theta \cdot \tan \theta \ge 1-\tan \theta .$$

Second solution. Set $x=\tan \alpha$, $y=\tan \beta$ and $z=\tan \gamma$. Then $x,y,z\ge 0$, $x+y+z\le 3$ and we have to prove that $$\cos 2\alpha +\cos 2\beta +\cos 2\gamma =\frac{1-x^2}{1+x^2}+\frac{1-y^2}{1+y^2}+\frac{1-z^2}{1+z^2}\ge 0.$$ This inequality can be written as $$\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+z^2}\ge \frac{3}{2}.$$ To prove the last inequality, it remains to use that $\frac{1}{1+t^2}\ge 1-\frac{t}{2}$ is equivalent to $t(t-1{)}^2\ge 0$.