No. This reads $${\mathbb E}[f(a,b)]\le f({\mathbb E}[a,b])$$ for an arbitrary discrete probability. This is true if and only if $f$ is a convex function. But $f(x,y)=\frac xy$ is not convex.

You might prefer the convex function $g(x,y)=\frac{x^2}y$, for which we thus have $${\mathbb E}\left[\frac{a^2}b\right]\le \frac{{\mathbb E}[a]^2}{{\mathbb E}[b]}.$$

No. This reads $${\mathbb E}[f(a,b)]\le f({\mathbb E}[a,b])$$ for an arbitrary discrete probability. This is true if and only if $f$ is a concave function (Jenssen Inequality). But $f(x,y)=\frac xy$ is not concave (nor convex).

You might prefer the convex function $g(x,y)=\frac{x^2}y$, for which we thus have $${\mathbb E}\left[\frac{a^2}b\right]\ge \frac{{\mathbb E}[a]^2}{{\mathbb E}[b]}.$$