Difference between revisions of "Aczel's Inequality"
(New page: '''Aczel's Inequality''' states that if <math>a_1^2>a_2^2+\cdots +a_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...) |
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| − | '''Aczel's Inequality''' states that if <math>a_1^2>a_2^2+\cdots +a_n^2</math>, then | + | '''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 |
<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 == | ||
| − | {{ | + | 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>. |
| + | |||
| + | <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>. | ||
== See also == | == See also == | ||
Revision as of 12:16, 30 January 2009
Aczel's Inequality states that if 
and 
, then
Proof
Let us get the function 
.
and since , then 
. Therefore, 
has to have at least one root, 
.
See also
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