Alternative proof for convexity of a set

Question

Consider a set $$ \mathcal{C} = \{(x, y)| a \leq x \leq b, y < 0, f(x) y \leq c\} $$ with $c > 0$ and the assumption that the reciprocal $\frac{1}{f(x)}$ is negative and convex for $a \leq x \leq b$. Convexity of this set is not too difficult to show, since the first two inequalities are not very troublesome and the third can be rearranged as $$ \frac{c}{f(x)}-y\leq 0, $$ where the l.h.s. is a convex function of $(x,y)$ for $a\leq x \leq b$ due to my previous assumptions about $\frac{1}{f(x)}$. Is there an alternative way to prove convexity of $\mathcal{C}$ without involving the reciprocal of $f(x)$, but rather only similar (implied) assumptions on $f(x)$? My motivation behind this is that I would like to investigate similar sets in higher dimensions, e.g., $y_1, y_2 <0$ and $f_1(x)y_1+f_2(x)y_2\leq c$ and check what conditions on $f_1(x), f_2(x)$ would be sufficient to produce a similarly convex set $\mathcal{C}$. Note that a simple rearrangement like $$ f(x)y-c \leq 0 $$ did not lead me to the desired solution, since the l.h.s is not a convex function of $(x,y)$ on the considered set. (I checked the Hessian of $f(x)y$, which is indefinite.)