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 $t(t-1)^2 \ge 0$.