Intuition behind reversal of Schwarz inequality for Minkowski space? I was recently working a problem which showed that, for a space with metric:
$ g = \begin{pmatrix} -1 & 0 \\ 0 & 1\end{pmatrix} $
The Schwarz inequality holds $ \textit{with the inequality reversed} $:
$ \left ( v_1 \cdot v_2 \right )^2 \geq v_1^2 v_2^2 $
I was able to solve the problem with quite a bit of algebra and reasoning, but I was left feeling very unenlightened. Is there any good intuition for this? Does it have anything to do with the fact that Minkowski space is a hyperbolic geometry?
Thanks in advance!
 A: The idea is to note that in Minkowski space
$$
v_1 \cdot v_2 = |v_1||v_2| \cosh\eta
$$
instead of the usual $v_1 \cdot v_2 = |v_1| |v_2| \cos \eta$. Then,
$$
(v_1\cdot v_2)^2 = v_1^2 v_2^2 \cosh^2 \eta \geq v_1^2 v_2^2
$$
since $\cosh \eta \geq 1$. 
EDIT: I jumped the gun a bit. Let $|v| = \sqrt{\pm v^2} \geq 0$ where the sign is positive if the vector $v$ is space-like and negative if the vector is time-like. 
The correct statement is that if both vectors are space-like or time-like, then
$$
(v_1 \cdot v_2)^2 = |v_1|^2 |v_2|^2 \cosh^2 \eta \geq |v_1|^2 |v_2|^2 = v_1^2 v_2^2 
$$
However, if one vector is space-like and the other is time-like, then
$$
(v_1 \cdot v_2)^2 = |v_1|^2 |v_2|^2 \sinh^2 \eta \geq  0 \geq - |v_1|^2 |v_2|^2 = v_1^2 v_2^2 
$$
A: The inequality direction is related to the convexity of the unit ball. In fact, the reverse Cauchy-Schwarz inequality holds also in Lorentz-Finsler theory in which the unit ball is no more a hyperboloid but it is still asymptotic to a (anisotropic) cone.
In short, if the origin of the vector space belongs to the convex set bounded by the unit sphere, as in Riemannian geometry, the direction of the inequality is the usual one. If the origin  stays outside the convex set bounded by the unit sphere then the direction is reversed. 
