# $a_ia_j+b_ib_j\leq \sqrt{a_i^2+b_i^2} \sqrt{a_j^2+b_j^2}$

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

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

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 $\sqrt{(u_i^2+u_j^2)(v_i^2+v_j^2)}\geq |u_iv_i+u_jv_j|$.