Consider the six dot product of four vectors $v_1,v_2,v_3,v_4$ on $\mathbb{R}^2$. Can all of them be negative? 
Consider the six dot products of four vectors $v_1,v_2,v_3,v_4$ on $\mathbb{R}^2$. Can all of them be negative?

If all dot products are negative, then the angle between each two vectors are larger than $\pi/2$. Intuitively, I don't think it's possible. But I don't know how to make a formal proof.
 A: Assume this is possible. Since you are only concerned with angles between the vectors (as you correctly note, the pairwise angles must be between $\pi/2$ and $3\pi/2$), you can rotate any such system of vectors so that $v_1$ is along the $x$-axis. Then $v_2$ must be in the 2nd or 3rd quadrant.
If $v_2$ is in the third, $v_3$ cannot be in the 4th and can only be in the second. If $v_2$ is in the second, $v_3$ cannot be in the 4th so must be in the 3rd.
In summary, there is no place to put $v_4$ so that it would be more than $\pi/2$ away from other placed vectors.
A: You can do this with simple algebra. Can you first convince yourself that there is no loss of generality taking $v_1 = (1,0)$?
Suppose all the inner products are negative. 
Write $v_2 = (a_2,b_2)$, $v_3 = (a_3,b_3)$, and $v_4 = (a_4,b_4)$. Compute the inner products with $v_1$ to find $a_2 < 0$, $a_3 < 0$, and $a_4 < 0$.
Since $a_2 a_3 + b_2 b_3 < 0$ and $a_2 a_3 > 0$ you get $b_2 b_3 < 0$. Likewise $b_2 b_4 < 0$ and $b_3 b_4 < 0$. This is impossible.
