Theorem: Cauchy-Schwarz Inequality

Let a_1,a_2,\dots,a_n and b_1,b_2,\dots,b_n be sequences of real numbers. Then we have

\left|\sum_{i=1}^na_ib_i\right|\leq\sqrt{\sum_{i=1}^na_i^2}\sqrt{\sum_{i=1}^nb_i^2}.

Proof : Consider the following quadratic function f:\mathbb{R}\rightarrow\mathbb{R}:

f(x)=\sum_{i=1}^n(a_ix+b_i)^2.

Note at this point that f(x)\geq0 for all x\in\mathbb{R}.

f(x)=\sum_{i=1}^n(a_i^2x^2+2a_ib_ix+b_i^2)

=\left(\sum_{i=1}^na_i^2\right)x^2+\left(2\sum_{i=1}^na_ib_i\right)x+\sum_{i=1}^nb_i^2.

That is f is a \bigcup or `+x^2‘ positive quadratic so has one or no roots. That means the roots are real and repeated or complex so that we have b^2-4ac\leq 0 where f(x)=ax^2+bx+c:

\left(2\sum_{i=1}^na_ib_i\right)^2-4\left(\sum_{i=1}^na_i^2\right)\left(\sum_{i=1}^nb_i^2\right)\leq0

\Rightarrow \left(\sum_{i=1}^na_ib_i\right)^2\leq \left(\sum_{i=1}^na_i^2\right)\left(\sum_{i=1}^nb_i^2\right)

Now take square roots (remembering \sqrt{x^2}=|x|.) \bullet

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