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


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


reference


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