Difference between revisions of "Aczel's Inequality"

Revision as of 15:05, 30 January 2009

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, $\Leftrightarrow$ $D=(a_1b_1-a_2b_2-\cdots -a_nb_n)^2- (a_1^2-a_2^2-\cdots -a_n^2)(b_1^2-b_2^2-\cdots -b_n^2)\geq 0$ .

Revision as of 13:16, 30 January 2009 (view source) BG Yoda (talk \| contribs) ← Older edit		Revision as of 15:05, 30 January 2009 (view source) BG Yoda (talk \| contribs) Newer edit →
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−	'''Aczel's Inequality''' states that if <math>a_1^2>a_2^2+\cdots +a_n^2</math> ~~and~~ <math>b_1^2>b_2^2+\cdots +b_n^2</math>, then	+	'''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

	<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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Difference between revisions of "Aczel's Inequality"

Revision as of 15:05, 30 January 2009

Proof

See also