Difference between revisions of "Aczel's Inequality"
Etmetalakret (talk | contribs) |
|||
(5 intermediate revisions by 4 users not shown) | |||
Line 1: | Line 1: | ||
− | ''' | + | '''Aczél's Inequality''' states that if <math>a_1^2>a_2^2+\cdots +a_n^2</math> or <math>b_1^2>b_2^2+\cdots +b_n^2</math>, then |
<center><math>(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).</math></center> | <center><math>(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).</math></center> | ||
+ | |||
== Proof == | == Proof == | ||
− | + | Consider the function <math>f(x)=(a_1 x - b_1)^2-\sum_{i=2}^n(a_i x - b_i)^2=</math> <math>(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)</math>. | |
+ | |||
+ | We have <math>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</math>, and from <math>a_1^2>a_2^2+\cdots +a_n^2</math> we get <math>\lim_{x\rightarrow \infty}f(x)\rightarrow \infty</math>. Therefore, <math>f(x)</math> must have at least one root, <math>\Leftrightarrow </math> <math>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</math>. | ||
+ | |||
+ | |||
+ | ==General Form== | ||
+ | Let <math>p_1,\dots,p_m \ge1 </math> such that <math>\sum_{i=1}^m\frac1{p_i} = 1</math> and let <center><math>(a_{11}, \dots,a_{1n}),</math></center><cmath>\vdots</cmath><center><math>(a_{m1}, \dots , a_{mn}) </math></center> | ||
+ | be <math>m</math> sequences of positive real numbers such that <math>a_{i1}^{ p_i} - a_{i2}^{ p_i} - \dots - a_{in}^{ p_i} > 0</math> for <math>i=1,\dots,m </math>. Then | ||
+ | |||
+ | <center><math> \prod_{i=1}^m a_{i1} - \prod_{i=1}^m a_{i2} -\dots- \prod_{i=1}^m a_{in} \ge\prod_{i=1}^m | ||
+ | (a_{i1}^{ p_i} - a_{i2}^{ p_i} - \dots - a_{in}^{ p_i})^\frac 1{ p_i}</math></center> | ||
+ | with equality if and only if all the sequences are proportional. | ||
+ | |||
+ | ==Examples== | ||
+ | |||
+ | '''Olympiad''' | ||
+ | Suppose <math>a_1, a_2, \ldots, a_n</math> and <math>b_1, b_2, \ldots, b_n</math> are real numbers such that <cmath> (a_1 ^ 2 + a_2 ^ 2 + \cdots + a_n ^ 2 -1)(b_1 ^ 2 + b_2 ^ 2 + \cdots + b_n ^ 2 - 1) > (a_1 b_1 + a_2 b_2 + \cdots + a_n b_n - 1)^2. </cmath> Prove that <math>a_1 ^ 2 + a_2 ^ 2 + \cdots + a_n ^ 2 > 1</math> and <math>b_1 ^ 2 + b_2 ^ 2 + \cdots + b_n ^ 2 > 1</math>. (USA TST 2004) | ||
− | + | == References == | |
+ | *Mascioni, Vania, [http://www.maths.soton.ac.uk/EMIS/journals/JIPAM/v3n5/045_02_www.pdf A note on Aczél-type inequalities], JIPAM volume 3 (2002), issue 5, article 69. | ||
+ | * Popoviciu, T., Sur quelques inégalités, Gaz. Mat. Fiz. Ser. A, 11 (64) (1959) 451–461 | ||
== See also == | == See also == | ||
* [[Inequalities]] | * [[Inequalities]] | ||
− | [[Category: | + | [[Category:Algebra]] |
− | [[Category: | + | [[Category:Inequalities]] |
{{stub}} | {{stub}} |
Latest revision as of 15:36, 29 December 2021
Aczél's Inequality states that if or , then
Contents
[hide]Proof
Consider the function .
We have , and from we get . Therefore, must have at least one root, .
General Form
Let such that and let
be sequences of positive real numbers such that for . Then
with equality if and only if all the sequences are proportional.
Examples
Olympiad Suppose and are real numbers such that Prove that and . (USA TST 2004)
References
- Mascioni, Vania, A note on Aczél-type inequalities, JIPAM volume 3 (2002), issue 5, article 69.
- Popoviciu, T., Sur quelques inégalités, Gaz. Mat. Fiz. Ser. A, 11 (64) (1959) 451–461
See also
This article is a stub. Help us out by expanding it.