\[\{ (a_i,b_i): i=1,...,n \}\]
\[ (\sum^n_{i=1}a^2_i)(\sum^n_{i=1}b^2_i) \ge (\sum^n_{i=1}a_ib_i)^2\]
.To prove it consider the quadratic
\[\begin{equation} \begin{aligned} q(x) &=(a_1x-b_1)^2+(a_2x-b_2)^2+...+(a_nx-b_n)^2 \\ &=(a^2_1+a_2^2+...+a^2_n)x^2-2(a_1b_1+a_2b_2+...+a_nb_n)x+(b^2_1+b^2_2+...+b^2_n ) \\ &= (\sum^n_{i=1} a^2_i)x^2-(2 \sum^n_{i=1} a_ib_i)x + (\sum^n_{i=1}b^2_i)\end{aligned} \end{equation}\]
.This quadratic has no roots, since it is the sum of positive terms, so always positive. The discriminant is less than zero therefore.
\[(-(2 \sum^n_{i=1} a_ib_i))^2-4(\sum^n_{i=1} a^2_i)(\sum^n_{i=1} b^2_i) \lt 0 \rightarrow (\sum^n_{i=1}a^2_i)(\sum^n_{i=1}b^2_i) \gt (\sum^n_{i=1}a_ib_i)^2\]