Intuition behind reversal of Schwarz inequality for Minkowski space?

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 $$ (v_1 \cdot v_2)^2 = |v_1|^2 |v_2|^2 \sinh^2 \eta \geq 0 \geq - |v_1|^2 |v_2|^2 = v_1^2 v_2^2 $$