Solution:
Let $a=\cos \alpha+i \sin \alpha$, $b=\cos \beta+i \sin \beta$, and $c=\cos \gamma+i \sin \gamma$.
We then have
$$a+b+c=1+i$$
where $a, b, c$ are complex numbers on the unit circle. Now, to minimize $\cos \alpha=\operatorname{Re}[a]$, consider a triangle with vertices $a, 1+i$, and the origin. We want $a$ as far away from $1+i$ as possible while maintaining a nonnegative imaginary part. This is achieved when $b$ and $c$ have the same argument, so $|b+c|=|1+i-a|=2$. Now $a, 0$, and $1+i$ form a $1-2-\sqrt{2}$ triangle. The value of $\cos \alpha$ is now the cosine of the angle between the 1 and $\sqrt{2}$ sides plus the $\frac{\pi}{4}$ angle from $1+i$. Call the first angle $\delta$. Then
$$\begin{aligned} \cos \delta & =\frac{1^{2}+(\sqrt{2})^{2}-2^{2}}{2 \cdot 1 \cdot \sqrt{2}} \\ & =\frac{-1}{2 \sqrt{2}} \end{aligned}$$
and
$$\begin{aligned} \cos \alpha & =\cos \left(\frac{\pi}{4}+\delta\right) \\ & =\cos \frac{\pi}{4} \cos \delta-\sin \frac{\pi}{4} \sin \delta \\ & =\frac{\sqrt{2}}{2} \cdot \frac{-1}{2 \sqrt{2}}-\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{7}}{2 \sqrt{2}} \\ & =\frac{-1-\sqrt{7}}{4} \end{aligned}$$