2
$\begingroup$

In Euclidean space ${\displaystyle \mathbb {R} ^{n}}$ with the standard inner product, the Cauchy–Schwarz inequality is

$$\left(\sum_{i=1}^n u_i v_i\right)^2\leq \left(\sum_{i=1}^n u_i^2\right) \left(\sum_{i=1}^n v_i^2\right)$$

but i can't see how they used in the following inequality:

$\forall (i,j)\in\{1,\ldots,n\}^{2}$: $$a_ia_j+b_ib_j\leq \sqrt{a_i^2+b_i^2} \sqrt{a_j^2+b_j^2} $$

  • What is u and what is v in that case could someone explain to me how : Cauchy–Schwarz inequality

reference

enter image description here

$\endgroup$
  • 6
    $\begingroup$ It seems to be a straightforward use of the inequality with $n=2$. So $u_1=a_i,u_2=a_j,v_1=b_i,v_2=b_j$. $\endgroup$ – almagest Jun 6 '16 at 12:08
1
$\begingroup$

Just $\sum\limits_{i=1}^nu_i^2\sum\limits_{i=1}^nv_i^2-\left(\sum\limits_{i=1}^nu_iv_i\right)^2=\sum\limits_{1\leq i<j\leq n}(u_iv_j-u_jv_i)^2\geq0$,

where $(u_iv_j-u_jv_i)^2\geq0$ it's $\sqrt{(u_i^2+u_j^2)(v_i^2+v_j^2)}\geq |u_iv_i+u_jv_j|$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.