The product of the roots is $\frac{5}{4}$. Use AM-GM to get that $1 = \frac{r_1}{2} + \frac{r_2}{4} + \frac{r_3}{5} + \frac{r_4}{8} \geq 4 \cdot \sqrt[4]{\frac{r_1}{2} \cdot \frac{r_2}{4} \cdot \frac{r_3}{5} \cdot \frac{r_4}{8}}=1$. Thus $\frac{r_1}{2} = \frac{r_2}{4} = \frac{r_3}{5} = \frac{r_4}{8}$ (equality only for equality in AM-GM) and the problem gets trivial.

additional condition that $ a,b,c$ are positive is missing,is it? Edited: yup,my mistake. I didn't note the inscription below "real".

Given that the roots are positive, Vieta's formula implies the positiveness of $ a,b,c$. Anyway, they are not required, are they?

By Vieta we get $r_{1}r_{2}r_{3}r_{4}=\frac{5}{4}$, now, by MG-MA one finds that \[\frac{r_{1}}{2}+\frac{r_{2}}{4}+\frac{r_{3}}{5}+\frac{r_{4}}{8}\geq 4\sqrt[4]{\frac{r_{1}r_{2}r_{3}r_{4}}{2\cdot 4\cdot 5\cdot 8}}.\]By replacing values in this expression we get \[1\geq 4\sqrt[4]{\frac{\frac{5}{4}}{2\cdot 4\cdot 5\cdot 8}}=1,\]Thus, $\frac{r_{1}}{2}=\frac{r_{2}}{4}=\frac{r_{3}}{5}=\frac{r_{4}}{8}$, and by this we get $r_{1}=\frac{1}{2}$, $r_{2}=1$, $r_{3}=\frac{5}{4}$, $r_{4}=2$, and we're done.