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 $a.$
I think this question will require using Vieta's Formulas, so I have
\begin{align*}
\sum_{i=1}^{4}r_i&=\frac{a}{4},\\
\sum_{1\leq i <j\leq 4}r_ir_j&=\frac{b}{4},\\
\sum_{1\leq i<j<k\leq 4}r_ir_jr_k&=\dfrac{c}{4},\\
\prod_{i=1}^{4}r_{i}&=\dfrac{5}{4}.\\
\end{align*}
However, I'm not sure if this helps, and if it does, how to use it. I also have $a,b,c>0$m but this doesn't seem to help at the moment. Any guidance would be greatly appreciated!!
Thanks in advance!!
Best Answer
Hint: Using $AM -GM$ inequality , we have:
$$\dfrac{\dfrac{r_1}{2} + \dfrac{r_2}{4} + \dfrac{r_3}{5} + \dfrac{r_4}{8}}{4} \geq \ (\dfrac{r_1}{2}\cdot\dfrac{r_2}{4}\cdot\dfrac{r_3}{5}\cdot \dfrac{r_4}{8})^{1/4}$$
$$\implies \dfrac{1}{4} \geq \dfrac{1}{4}$$
As AM = GM, thus $\dfrac{r_1}{2} = \dfrac{r_2}{4} = \dfrac{r_3}{5} = \dfrac{r_4}{8}$
Can you proceed further from here?