How prove this $\overrightarrow{r_{i}}\cdot\overrightarrow{r_{j}}\ge\frac{1}{2}$ let $a,b,c$ are real numbers,and such $a+b+c=0,a^2+b^2+c^2=1$, we define:
$\overrightarrow{r}=(x_{i},y_{i},z_{i})(i=1,2,3,4,5,6)$,where $\{x_{i},y_{i},z_{i}\}=\{a,b,c\}$,

show that: there are exst $\overrightarrow{r_{i}}\neq\overrightarrow{r_{j}}$,such $$\overrightarrow{r_{i}}\cdot\overrightarrow{r_{j}}\ge\dfrac{1}{2}$$

My try: $$a+b+c=0,\Longrightarrow c=-a-b$$
then $$a^2+b^2+c^2=1\Longrightarrow a^2+b^2+(a+b)^2=1\Longrightarrow a^2+b^2+ab=\dfrac{1}{2}$$
let $$a=x+y,b=x-y\Longrightarrow (x+y)^2+(x-y)^2+x^2-y^2=\dfrac{1}{2}$$
$$\Longrightarrow 3x^2+y^2=\dfrac{1}{2}$$
then I can't ,Thank you

The points $r_i$ lie on a unit circle (the intersection of the plane $x+y+z=0$ and the sphere $x^2 + y^2 + z^2 = 1$). There isn't enough "room" on a circle for 6 points that are all more than 60 degrees apart. So, there must be two of them (say $r_i$ and $r_j$) whose angles differ by 60 degrees or less. Then $\vec{r_i} \cdot \vec{r_j} \ge \cos 60^\circ = \tfrac12$.
