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

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"
Art of Problem Solving's olympiad training program WOOT starts on September 8. Train with the top high school students in the the world! Click here to enroll today!
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us