## 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 $\sqrt{x^2}=|x|$.) $\bullet$