1985 OIM Problems/Problem 3

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Problem

Find the roots $r_1$, $r_2$, $r_3$, and $r_4$ of the equation: \[4x^4-ax^3+bx^2-cx+5=0\] knowing that they're all real, positives and that: \[\frac{r_1}{2}+\frac{r_2}{4}+\frac{r_3}{5}+\frac{r_4}{8}=1\]

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

By Vieta's, $r_1r_2r_3r_4=\frac54$. Because the roots are real and positive, by AM-GM, $\frac{r_1}{2}+\frac{r_2}{4}+\frac{r_3}{5}+\frac{r_4}{8}\ge4\sqrt[4]{r_1r_2r_3r_4\frac{1}{5(64)}}=1$, so by the equality condition $\frac{r_1}{2}=\frac{r_2}{4}=\frac{r_3}{5}=\frac{r_4}{8}=\frac14$, so $(r_1,r_2,r_3,r_4)=\boxed{(\frac12,1,\frac54,2)}$.

See also

https://www.oma.org.ar/enunciados/ibe1.htm