Cauchy-Schwarz Inequality
==
- For $x=(x_1,\dots,x_n),y=(y_1,\dots,y_n)\in \mathbb{R}^n$, denote
\begin{align}
\Vert x\Vert &= \sqrt{x_1^2+\dots+x_n^2}\\
x\cdot y&= x_1y_1+\cdots+x_ny_n
\end{align}
- The question in concern is how to prove the following inequality:
$$
\Vert x\Vert^2\Vert y \Vert^2\geq (x\cdot y)^2.
$$
> Proof. Let $$z = \left(\frac{x\cdot y}{\Vert y\Vert}\right)\frac{y}{\Vert y\Vert}.$$
> Since $$(x-z)\cdot z=x\cdot z+z\cdot z=\frac{(x\cdot y)^2}{\Vert y\Vert^2}-\frac{(x\cdot y)^2}{\Vert y\Vert^2}=0,$$
> it follows that
> \begin{align}
> \Vert x\Vert^2&= x\cdot x = [(x-z)+z]\cdot [(x-z)+z]\\
> &=\Vert x-z\Vert^2+\Vert z\Vert^2\geq \Vert z\Vert^2=\frac{(x\cdot y)^2}{\Vert y\Vert^2},
> \end{align}
> and consequently,
> $$\Vert x\Vert^2\Vert y\Vert^2\geq (x\cdot y)^2.$$

In what follows, we are going to give an alternative proof.
> **Alternative proof.**
> Let $t\in \mathbb{R}$. Then
> \begin{align}
> 0&\leq (tx+y)\cdot (tx+y)=\Vert x\Vert^2t^2+2(x\cdot y) t+\Vert y \Vert^2.
> \end{align}
> This implies that
> $$
> [2(x\cdot y)]^2-4\Vert x\Vert^2\Vert y \Vert^2\leq 0
> $$
> so that
> $$
> (x\cdot y)^2\leq \Vert x\Vert^2\Vert y\Vert^2.
> $$

## Some more examples
Consider an closed interval $[a,b]$. Prove that for any two continuous functions $f(x)$ and $g(x)$, $$\left(\int_a^b[f(x)]^2dx\right)\left(\int_a^b[g(x)]^2dx\right)\geq \left(\int_a^bf(x)g(x)dx\right)^2.$$
> **Proof.** For any contionous function $f(x)$, we denote
$$\Vert f\Vert = \left(\int_a^b [f(x)]^2dx\right)^{1/2}.$$
Let $$\lambda = \frac{\int_a^bf(x)g(x)dx}{\Vert g\Vert^2}.$$
Then
\begin{align}
\int_a^b [f(x)-\lambda g(x)]g(x)dx&=\int_a^bf(x)g(x)dx-\lambda\Vert g\Vert^2=0.
\end{align} Hence,
\begin{align}
\int_a^b[f(x)]^2dx&=\int_a^b[(f(x)-\lambda g(x))+\lambda g(x)]^2dx\\
&=\int_a^b[f(x)-\lambda g(x)]^2dx+2\lambda\int_a^b[f(x)-\lambda g(x)]g(x)dx+\lambda2\int_a^b[g(x)]^2dx\\
&\geq \lambda^2\int_a^b[g(x)]^2dx
\end{align}