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Difference between revisions of "Aczel's Inequality"
 
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'''Aczel'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
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'''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>
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== Proof ==
 
== Proof ==
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Let us get 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>.
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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>.
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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>.
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==General Form==
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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>
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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 
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<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
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(a_{i1}^{ p_i} - a_{i2}^{ p_i} - \dots - a_{in}^{ p_i})^\frac 1{ p_i}</math></center>
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with equality if and only if all the sequences are proportional.
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==Examples==
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'''Olympiad'''
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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)
  
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<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 since <math>a_1^2>a_2^2+\cdots +a_n^2</math>, then <math>\lim_{x\rightarrow \infty}f(x)\rightarrow \infty</math>. Therefore, <math>f(x)</math> has to 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>.
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== References ==
  
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*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.
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* Popoviciu, T., Sur quelques inégalités, Gaz. Mat. Fiz. Ser. A, 11 (64) (1959) 451–461
 
== See also ==
 
== See also ==
 
* [[Inequalities]]
 
* [[Inequalities]]
  
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[[Category:Inequality]]
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[[Category:Algebra]]
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[[Category:Theorems]]
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[[Category:Inequalities]]
 
{{stub}}
 
{{stub}}

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

See also

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