Inspired by Calvin Lin

Since $r_1$ , $r_2$ , $r_3$ and $r_4$ are positive by AM-GM inequality, we have:

$\quad \dfrac {r_1}{2} + \dfrac {r_2}{4} + \dfrac {r_3}{5} + \dfrac {r_4}{8} \ge 4 \sqrt [4] {\dfrac {r_1}{2} \dot{} \dfrac {r_2}{4} \dot{} \dfrac {r_3}{5} \dot{} \dfrac {r_4}{8}}$

And by Vieta formulas, we have: $\space r_1r_2r_3r_4 = \dfrac {5}{4}$

$\quad \Rightarrow \dfrac {r_1}{2} + \dfrac {r_2}{4} + \dfrac {r_3}{5} + \dfrac {r_4}{8} \ge 4 \sqrt [4] {\dfrac {1}{2 \dot{} 4 \dot{} 5 \dot{} 8} \dot{} \dfrac {5}{4}} = 1$

This means that equality happens, therefore,

$\quad \Rightarrow \dfrac {r_1}{2} = \dfrac {r_2}{4} = \dfrac {r_3}{5} = \dfrac {r_4}{8} = \dfrac {1}{4}$

By Vieta formulas again, we have:

$\begin{aligned} \quad \Rightarrow \dfrac {a}{4} & = r_1+r_2+r_3+r_4 \\ & = \dfrac {2}{4} + \dfrac {4}{4} + \dfrac {5}{4} + \dfrac {8}{4} \\ \Rightarrow a & = 2 + 4 + 5 + 8 = \boxed{19} \end{aligned}$