Do you really love polynomials

From Vieta's Formula and the given conditions, the problem can be restated as follow:

$r_1, r_2, r_3, r_4$ are positive real numbers such that $r_1r_2r_3r_4 = \frac{5}{4}$ and $\frac{r_1}{2}+\frac{r_2}{4}+\frac{r_3}{5}+\frac{r_4}{8} = 1$ .

Find $a = 4(r_1+r_2+r_3+r_4)$ .

Applying the AM-GM inequality, we have

$\frac{\frac{r_1}{2}+\frac{r_2}{4}+\frac{r_3}{5}+\frac{r_4}{8}}{4} \geq \sqrt[4]{\frac{r_1}{2}\cdot\frac{r_2}{4}\cdot\frac{r_3}{5}\cdot\frac{r_4}{8}}$

Substituting the known values of the expressions, we have

$\frac{1}{4} \geq \sqrt[4]{\frac{\frac{5}{4}}{(2)(4)(5)(8)}}$

$\frac{1}{4} \geq \sqrt[4]{\frac{1}{2^8}}$

$\frac{1}{4} \geq \frac{1}{4}$

The equality holds if and only if $\frac{r_1}{2} = \frac{r_2}{4} = \frac{r_3}{5} = \frac{r_4}{8}$ .

So $\frac{r_1}{2} = \frac{r_2}{4} = \frac{r_3}{5} = \frac{r_4}{8} = \frac{1}{4}$ .

$r_1 = \frac{1}{2}, r_2 = 1, r_3 = \frac{5}{4}, r_4 = 2$ .

Hence, $a = 4(r_1+r_2+r_3+r_4) = 4(\frac{1}{2}+1+\frac{5}{4}+2) = 19$ .