Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1991 APMO Problems/Problem 3"

(Created page with "== Problem == Let <math>a_1</math>, <math>a_2</math>, <math>\cdots</math>, <math>a_n</math>, <math>b_1</math>, <math>b_2</math>, <math>\cdots</math>, <math>b_n</math> be positive...")
(No difference)

Revision as of 17:40, 4 August 2011

Problem

Let $a_1$, $a_2$, $\cdots$, $a_n$, $b_1$, $b_2$, $\cdots$, $b_n$ be positive real numbers such that $a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_n$. Show that \[\frac{a_1^2}{a_1 + b_1} + \frac{a_2^2}{a_2 + b_2} + \cdots + \frac{a_n^2}{a_n + b_n} \geq \frac{a_1 + a_2 + \cdots + a_n}{2}\]

Solution

By Cauchy-Schwarz, $\left(\sum \dfrac{a_i^2}{a_i+b_i}\right)\left(\sum a_i+b_i\right)\geq \left(\sum a_i\right) ^2$, so $\sum \dfrac{a_i^2}{a_i+b_i}\geq \dfrac{\left(\sum a_i\right) ^2}{\sum (a_i+b_i)}=\dfrac{\sum a_i}{2} since$\sum a_i=\sum b_i$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1991_APMO_Problems/Problem_3&oldid=41187"