Aczel's Inequality

Aczel's Inequality states that if $a_1^2>a_2^2+\cdots +a_n^2$ and $b_1^2>b_2^2+\cdots +b_n^2$ , 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, $\Leftrightarrow$ $D=(a_1b_1-a_2b_2-\cdots -a_nb_n)^2- (a_1^2-a_2^2-\cdots -a_n^2)(b_1^2-b_2^2-\cdots -b_n^2)\geq 0$ .

Page

Toolbox

Search

Aczel's Inequality

Proof

See also