Minimum value of $(\cos\theta_1+\cos\theta_2+\cos\theta_3)$ If $\hat{x}$, $\hat{y}$ and $\hat{z}$ are three unit vectors in three-dimensional space, then the minimum value of\
$|\hat{x}+\hat{y}|^2+|\hat{y}+\hat{z}|^2+|\hat{z}+\hat{x}|^2$ is:
My Attempt : Let $\theta_1$ be angle between $\hat{x}$ and $\hat{y}$, $\theta_2$ be angle between $\hat{y}$ and $\hat{z}$ and $\theta_3$ be angle between $\hat{z}$ and $\hat{x}$.
$\therefore \theta_1,\theta_2$ and $\theta_3$ are the angles between any 2 edges of a tetrahedron from a single vertex.
$\therefore |\hat{x}+\hat{y}|^2+|\hat{y}+\hat{z}|^2+|\hat{z}+\hat{x}|^2=(2+2\cos\theta_1)+(2+2\cos\theta_2)+(2+2\cos\theta_3)$.
$=6+2(\cos\theta_1+\cos\theta_2+\cos\theta_3)$
Can anyone please tell me what to do next?
 A: You can rotate the picture and assume $\hat{x} = (\cos \theta, \sin \theta, 0)$ and $\hat{y} = (\cos \theta, -\sin \theta, 0)$ for some $0 \leq \theta \leq {\pi \over 2}$. Writing $z = (z_1,z_2,z_3)$ you have
$$\hat{x} \cdot \hat{y} + \hat{x} \cdot \hat{z} + \hat{y} \cdot \hat{z} = \cos^2 \theta - \sin^2 \theta + 2z_1\cos \theta $$
This is minimized for $z = (-1,0,0)$, in which case we have
$$\hat{x} \cdot \hat{y} + \hat{x} \cdot \hat{z} + \hat{y} \cdot \hat{z} = \cos^2 \theta - \sin^2 \theta - 2\cos \theta $$
$$= 2\cos^2 \theta - 2\cos \theta - 1$$
The function $2x^2 - 2x - 1$ has its minimum at $x = {1 \over 2}$, so $2\cos^2 \theta - 2\cos \theta - 1$ is minimized when $\theta = {\pi \over 3}$.
Thus we have $\hat{x} = ({1 \over 2}, {\sqrt{3} \over 2}, 0)$, $\hat{y} = ({1 \over 2}, -{\sqrt{3} \over 2}, 0)$, and $z = (-1,0,0)$, corresponding to 
$\hat{x} \cdot \hat{y} + \hat{x} \cdot \hat{z} + \hat{y} \cdot \hat{z} = 2({1 \over 2})^2  - 2{1 \over 2}  + 1 = -{3 \over 2}$, so that
$$|\hat{x}+\hat{y}|^2+|\hat{y}+\hat{z}|^2+|\hat{z}+\hat{x}|^2 \geq 6 - 3$$
$$ = 3$$
A: Here's a hint, you're missing a constraint.  We know that if two of the vectors are the same, and one is 180 degrees in direction, we can get to 4.  E.g., $(1,0,0),$ $(0,1,0),$ $(-1,0,0).$
Try to constrain $\theta_i$ in some way to see if you can beat that value.
A: $$\sum |\vec{x}+\vec{y}|^2=6+2\sum \vec{x}\cdot \vec{y}\geq 6-3\geq 3$$
Above  i am using 
$$|\vec{x}+\vec{y}+\vec{z}|^2\geq 0\Longrightarrow 3+2\sum \vec{x}\cdot \vec{y}\geq 0$$
$$\Longrightarrow \sum \vec{x}\cdot \vec{y}\geq -\frac{3}{2}$$
