Difference between revisions of "Aczel's Inequality"

Latest revision as of 15:36, 29 December 2021

Aczél'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

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

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

General Form

Let $p_1,\dots,p_m \ge1$ such that $\sum_{i=1}^m\frac1{p_i} = 1$ and let

$(a_{11}, \dots,a_{1n}),$

$\vdots$

$(a_{m1}, \dots , a_{mn})$

be $m$ sequences of positive real numbers such that $a_{i1}^{ p_i} - a_{i2}^{ p_i} - \dots - a_{in}^{ p_i} > 0$ for $i=1,\dots,m$ . Then

$\prod_{i=1}^m a_{i1} - \prod_{i=1}^m a_{i2} -\dots- \prod_{i=1}^m a_{in} \ge\prod_{i=1}^m (a_{i1}^{ p_i} - a_{i2}^{ p_i} - \dots - a_{in}^{ p_i})^\frac 1{ p_i}$

with equality if and only if all the sequences are proportional.

Examples

Olympiad Suppose $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ are real numbers such that $(a_1 ^ 2 + a_2 ^ 2 + \cdots + a_n ^ 2 -1)(b_1 ^ 2 + b_2 ^ 2 + \cdots + b_n ^ 2 - 1) > (a_1 b_1 + a_2 b_2 + \cdots + a_n b_n - 1)^2.$ Prove that $a_1 ^ 2 + a_2 ^ 2 + \cdots + a_n ^ 2 > 1$ and $b_1 ^ 2 + b_2 ^ 2 + \cdots + b_n ^ 2 > 1$ . (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

@@ Line 14: / Line 14: @@
 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
+<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} \le \prod_{i=1}^m a_{i1} - \prod_{i=1}^m a_{i2} -\dots- \prod_{i=1}^m a_{in} </math></center>
+(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 ==
@@ Line 26: / Line 30: @@
 * [[Inequalities]]
-[[Category:Inequality]]
+[[Category:Algebra]]
-[[Category:Theorems]]
+[[Category:Inequalities]]
 {{stub}}

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