for any $x>0$, $x \in \mathbb{R}^n$, whenever $a^T x \leq 1$, we have $b^T x \leq 1$; does this imply $a \leq b$?

Question

given $a>0$, $a$ is fixed, $a,b,x \in \mathbb{R}^n$, for any $x>0$, if whenever $a^T x \leq 1$ we have $b^T x \leq 1$, then is it true that $a \geq b$?

Answer 1

It actually implies $a \geq b.$ We can take a limit of vectors $x$ with all positive components, approaching $$ \left( \frac{1}{a_1},0,0, \ldots,0 \right) $$ where $a_1 > 0.$

But then $$ \frac{b_1}{a_1} \leq 1, $$ with $a_1$ positive $$ b_1 \leq a_1.$$ It is possible to have $b_1$ negative or $0$

Same for each $b_i$