Proving an inequality involving dot products

inequalitylinear algebra

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 $x_3 y_1 > 0$.

  • Let $ \arg(x_1) = \alpha, \arg(x_2) = \beta, \arg(x_3) = \gamma$.Then these are all (strictly) between $0^\circ$ and $90^\circ$.
  • Let $ \arg(y_1) = p, \arg(y_2) = q, \arg(y_1 + y_2) = r$, where $r$ is in-between $p$ and $q$.
  • Since $x_1 y_2, x_1 y_2 > 0$, so $ \alpha - 90^\circ < p, q < \alpha + 90^\circ$. In particular, $ p, q \not \in [180^\circ, 270^\circ]$, which allows us to do the following arithmetic (instead of having to consider when angles wrap around).
  • Since $x_2 \cdot (y_1 + y_2) < 0$ (resp $x_3 \cdot (y_1 + y_2) < 0$), so $r \not \in [\beta - 90^\circ, \beta + 90^\circ]$. (resp $r \not \in [\gamma - 90^\circ, \gamma + 90^\circ]$).
  • Since $x_2 y_2 > 0$, so $q \in [ \beta - 90^\circ, \beta + 90^\circ]$.
  • Since $x_3 y_1 > 0$, so $ p \in [\gamma - 90^\circ, \gamma + 90^\circ]$.
  • However, this contradicts that $r$ is in-between $p$ and $q$.

In higher dimensions, I believe that there is enough degrees of freedom to work around these constraints. Try running several numerical simulations.

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