Prove that if $0<x\le y$ then $1/y \le 1/x$ using ordered field axioms

Question

Suppose that $x,y,z \in \mathbb{R}^+$.

Then, either $x = y$ or $x \leq y$.

If $x = y$, then $x^{-1} = y^{-1}$.

If $x < y$, then via proof by contradiction let's suppose $x < y$ implies $x^{-1} < y^{-1}$.

$$(x)(x^{-1})<(x)(y^{-1})$$ and hence $$1<(x)y^{-1}$$ implies $$y<x$$

How should I take it further?

Answer 1

Suppose that $0 < x \leq y$. Then by the properties of inequalities, and the fact that $x^{-1}, y^{-1}$ exist in said field, $0< \frac{x}{y} \leq 1$. Further, $ 0 < \frac{1}{y} \leq \frac{1}{x} $.

To elaborate further, we were able to say that $x^{-1}*0=0$ and $y^{-1}*0=$ since we are in a field.