Smallest infinite place of a rational function field

Let $$f$$ be a separable irreducible polynomial of degree $$n$$ with $$f \in F[x]$$, where $$F$$ is a function field. I try to understand the following part of an article:

..., we use the minimum infinite valuation of the roots $$\alpha_i$$ of $$f$$. When $$F$$ is a rational function field, the infinite valuation is the negative of the degree of a root. We compute these valuation at all infinite places of F and call the smallest one $$\mathit{v}_0$$.

Hence $$\mathit{v}_\infty(\alpha_i) \geq {\mathit{v}}_{0}$$ and $$\mathit{v}_0 \leq 0$$.

As it is written, if $$F$$ is a rational function field the infinite valuation is the negative of the degree. Additionally there is onyl one place at infinity, let's say $$\mathfrak{p}_\infty$$, so $$\mathit{v}_0 = \mathfrak{p}_\infty$$. However I am confused by the meaning of that.

What is the meaning of $$\mathfrak{p}_\infty \leq 0$$ and why should $$\mathit{v}_\infty(\alpha_i) = -deg(\alpha_i) \geq \mathfrak{p}_\infty$$?

Is my understanding of $$\mathit{v}_0$$ wrong? Right now the two inequalities look like a contradiction for me.