$(a)$ $x\leq g(xy)\implies x\leq f(t)+\frac{xy}{t}$, for every $t>0$. Letting $t=2y$, we recover that $x/2 \leq f(2y)$, hence the claim. $(b)$ For this part, we want to show that, $x/2 \leq g(xy)$, or equivalently, $\inf_{t>0}\{f(t)+\frac{xy}{t}\}\geq \frac{x}{2}$. Hence, it suffices to show $$ f(t)+\frac{xy}{t}\geq \frac{x}{2}, \forall t>0. $$If $t\leq y$, the term, $\frac{xy}{t}\geq x \geq \frac{x}{2}$, and since $f(\cdot)$ is non-negative, we are done. If $t>y$, then using the fact that $f(\cdot)$ is increasing, $f(t)\geq f(y)\geq x$, hence we get the same domination. $\square$