Polynomial Positive Root Condition
The polynomial $ 4x^4 - ax^3 + bx^2 - cx + 5, $ where $ a, $ $ b, $ and $ c $ are real coefficients, has four positive real roots $ r_1, $ $ r_2, $ $ r_3, $ $ r_4, $ such that
\[\frac{r_1}{2} + \frac{r_2}{4} + \frac{r_3}{5} + \frac{r_4}{8} = 1.\]Find $ a $.
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- $\frac{a}{b}$
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Solution
By AM-GM,
\begin{align*}
\frac{r_1}{2} + \frac{r_2}{4} + \frac{r_3}{5} + \frac{r_4}{8} &\ge 4 \sqrt[4]{\frac{r_1}{2} \cdot \frac{r_2}{4} \cdot \frac{r_3}{5} \cdot \frac{r_4}{8}} \\
&= 4 \sqrt[4]{\frac{r_1 r_2 r_3 r_4}{320}}.\end{align*}Since $ \frac{r_1}{2} + \frac{r_2}{4} + \frac{r_3}{5} + \frac{r_4}{8} = 1, $ this gives us
\[r_1 r_2 r_3 r_4 \le \frac{320}{4^4} = \frac{5}{4}.\]By Vieta's formulas, $ r_1 r_2 r_3 r_4 = \frac{5}{4}, $ so by the equality condition in AM-GM,
\[\frac{r_1}{2} = \frac{r_2}{4} = \frac{r_3}{5} = \frac{r_4}{8} = \frac{1}{4}.\]Then $ r_1 = \frac{4}{2} = \frac{1}{2}, $ $ r_2 = 1, $ $ r_3 = \frac{5}{4}, $ and $ r_4 = 2, $ so
\[r_1 + r_2 + r_3 + r_4 = \frac{1}{2} + 1 + \frac{5}{4} + 2 = \frac{19}{4}.\]So by Vieta's formulas, $ a = \boxed{19} $.
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