AoPSWiki
Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.

Aczel's Inequality

From AoPSWiki

Aczel's Inequality states that if a_1^2>a_2^2+\cdots +a_n^2 or b_1^2>b_2^2+\cdots +b_n^2, then

(a_1b_1-a_2b_2-\cdots -a_nb_n)^2\geq (a_1^2-a_2^2-\cdots -a_n^2)(b_1^2-b_2^2-\cdots -b_n^2).

Proof

Let us get the function f(x)=(a_1 x - b_1)^2-\sum_{i=2}^n(a_i x - b_i)^2= (a_1^2-a_2^2-\cdots -a_n^2)x^2-2(a_1b_1-a_2b_2-\cdots -a_nb_n)x+(b_1^2-b_2^2-\cdots -b_n^2).

f\left( \frac{b_1}{a_1} \right)=-\sum_{i=2}^n\left(a_i \frac{b_1}{a_1} - b_i\right)^2\leq 0 and since a_1^2>a_2^2+\cdots +a_n^2, then \lim_{x\rightarrow \infty}f(x)\rightarrow \infty. Therefore, f(x) has to have at least one root, \Leftrightarrow D=(a_1b_1-a_2b_2-\cdots -a_nb_n)^2- (a_1^2-a_2^2-\cdots -a_n^2)(b_1^2-b_2^2-\cdots -b_n^2)\geq 0.

See also

This article is a stub. Help us out by expanding it.

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/Aczel%27s_Inequality"
Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us