We want to show that $x < y \Rightarrow F(x) \leq F(y)$, i.e. $$\frac{1}{x}\int_0^x f(t)\,\mathrm{d}t \leq \frac{1}{y}\int_0^y f(t)\,\mathrm{d}t$$ We note that $$\frac{1}{y}\int_0^y f(t)\,\mathrm{d}t = \frac{1}{y}\left[\int_0^x f(t)\,\mathrm{d}t+\int_x^y f(t)\,\mathrm{d}t\right]$$ Therefore, we can rewrite our inequality as $$\left(\frac{y}{x}-1\right)\int_0^x f(t)\,\mathrm{d}t \leq \int_x^y f(t)\,\mathrm{d}t$$ We multiply both sides by $\frac{1}{y-x}$ to get $$\frac{\frac{y}{x}-1}{y-x}\int_0^x f(t)\,\mathrm{d}t = \frac{1}{x}\int_0^x f(t)\,\mathrm{d}t \leq \frac{1}{y-x}\int_x^y f(t)\,\mathrm{d}t$$ The left-hand side is the average value of $f(t)$ on $(0, x)$, while the right-hand side is the average value of $f(t)$ on $(x, y)$. The left-hand side is therefore bounded above by $\sup_{t\in (0, x)} f(t)$, while the right-hand side is bounded below by $\inf_{t\in (x, y)} f(t)$. As $f$ is increasing, we therefore have that $$\frac{1}{x}\int_0^x f(t)\,\mathrm{d}t \leq \sup_{t\in (0, x)} f(t) \leq \inf_{t\in (x, y)} f(t) \leq \frac{1}{y-x}\int_x^y f(t)\,\mathrm{d}t$$ so we are done.