Here, $\cdot$ is the dot product. Suppose we have $x_1,x_2,x_3,y_1,y_2\in\mathbb R^2$. There is a system of inequalities of the following:
$$x_1\cdot y_1>0\\x_1\cdot y_2>0\\x_2\cdot y_2>0\\x_2\cdot y_1+x_2\cdot y_2<0\\x_3\cdot y_1+x_3\cdot y_2<0\\x_1>0,~x_2>0,~x_3>0.$$
Here, $x_i>0$ means all the elements of $x_i$ are strictly positive. When the inequalities hold true, can we show the following?$$ x_3\cdot y_1\leq 0$$
I ran some numerical simulations and found it should always hold.. but I cannot show why it should be so.
Furthermore, in general, if $x_1,x_2,x_3,y_1,y_2\in \mathbb R^n$ for $n\geq 2$, does the inequality still hold true?
Any help will be really appreciated.
Best Answer
In 2 dimensions, you can prove it by considering the angles to the x-axis (taken counter clockwise). We restrict the range of these angles to be $(-90^\circ, 270^\circ]$.
Proof by contradiction.
Suppose $x_3 y_1 > 0$.
In higher dimensions, I believe that there is enough degrees of freedom to work around these constraints. Try running several numerical simulations.