Cauchy-schwarz inequality

Consider the quadratic, $(a_1x+b_1)^2+(a_2x+b_2)^2+...(a_nx+b_n)^2 = 0$ . Expanding, we find the equation to be of the form $Ax^2+Bx+C$ , where $A=\sum_{i=1}^n a_i^2$ , $B=2\sum_{j=1}^n a_jb_j$ , and $C=\sum_{k=1}^n b_k^2.$ The equation will have a solution when the discriminant is greater than or equal to 0, so $B^2-4AC \geq 0$ . Substituting the above values of A, B, and C leaves us with the Cauchy-Schwarz Inequality, which states that $(a_1b_1+a_2b_2+...+a_nb_n)^2 \geq (a_1^2+a_2^2+...+a_n^2)(b_1^2+b_2^2+...+b_n^2)$ , or, in the more compact sigma notation, $\left(\sum a_ib_i\right) \geq \left(\sum a_i^2\right)\left(\sum b_i^2\right)$

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Cauchy-schwarz inequality