Difference between revisions of "1985 OIM Problems/Problem 3"
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== Solution == | == Solution == | ||
− | {{ | + | By Vieta's, <math>r_1r_2r_3r_4=\frac54</math>. Because the roots are real and positive, by AM-GM, <math>\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 </math>\frac{r_1}{2}=\frac{r_2}{4}=\frac{r_3}{5}=\frac{r_4}{8}=\frac14<math>, so </math>(r_1,r_2,r_3,r_4)=\boxed{(\frac12,1,\frac54,2)}$. |
== See also == | == See also == | ||
https://www.oma.org.ar/enunciados/ibe1.htm | https://www.oma.org.ar/enunciados/ibe1.htm |
Revision as of 22:46, 8 April 2024
Problem
Find the roots , , , and of the equation: knowing that they're all real, positives and that:
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
By Vieta's, . Because the roots are real and positive, by AM-GM, \frac{r_1}{2}=\frac{r_2}{4}=\frac{r_3}{5}=\frac{r_4}{8}=\frac14(r_1,r_2,r_3,r_4)=\boxed{(\frac12,1,\frac54,2)}$.