Prove that $-3/2\leq\cos a + \cos b + \cos c\leq 3$? 
Given 3 non-null vectors $v,u,w$ and angles $a=(u,v), b=(u,w), c=(v,w)$, Prove that $-3/2\leq\cos a + \cos b + \cos c\leq 3$.

I've managed to prove that: $\cos a + \cos b + \cos c\leq 3$ basically arguing that $\cos \theta$ is bounded by $-1,1$ using the inequality of Cauchy-Schwarz. I reasoned that as the maximum value of $\cos x$ is $1$, we could have three angles such that $1+1+1=3$ which is the right answer, but I guess I reasoned incorrectly. And for the lower bound:
$$-3/2\leq\cos a + \cos b + \cos c$$
I have no idea on what to do.

The picture is basically dividing a circle into 3 sectors and $a,b,c$ are the angles of each sector. 
Since, $c=2\pi-a-b$, $\cos c= \cos (a+b)$ and
$$\cos a + \cos b + \cos c = \cos a +\cos b + \cos a \cos b - \sin a \sin b $$
writing $x= \cos a,\; y= \cos b$, and  $\sin a =\sqrt{1-x^2} $, etc. (We can always choose $+\sqrt{}$ by choosing the angles to be <$\pi$) We have,
$$ x+y+xy -\sqrt{1-x^2}\sqrt{1-y^2}, \quad x,y\in[-1,1] $$
By A.M. > G.M., 
$$-\sqrt{1-x^2}\sqrt{1-y^2} \ge -\frac{2-x^2-y^2}{2} =-1 +\frac{x^2+y^2}{2}$$
Thus,
\begin{align} x+y+xy -\sqrt{1-x^2}\sqrt{1-y^2} & \ge x+y+xy-1+ \frac{x^2+y^2}{2}\\
 &= \frac{1}{2}(x+y+1)^2 -\frac{3}{2} \ge -\frac{3}{2}.
 \end{align}
And equality holds if and only if $x=y=-1/2$, which means the angles are all $ \frac{2\pi}{3}$.
