Cauchy-Schwarz Inequality Proof

Here is the simple proof of Cauchy Schwarz Inequality I found that is easier to understand than other resources on the internet

Here we first start off drafting few statements we already know

$$\vec{b}=\overrightarrow{b_{\perp }}+\overrightarrow{b_{\parallel }}$$

1. Any vector can be broken down into two vectors: one that is parallel to particular subspace and one that is perpendicular to the subspace

$$\overrightarrow{b_{\parallel }}=\frac{\vec{d}}{\Vert d\Vert }\frac{{\vec{b}}^T\vec{d}}{\Vert d\Vert }$$

2. The projection vector onto a vector d is given as above.

$${\overrightarrow{b_{\perp }}}^2=(\vec{b}-\overrightarrow{b_{\parallel }}{)}^2\ge 0$$

3. Lastly, the magnitude of a perpendicular vector must be equal or greater than 0.

With that being said, we can prove this Inequality

$$\begin{array}{r}(\vec{b}-\overrightarrow{b_{\parallel }}{)}^2\ge 0\\ (\vec{b}-\frac{\vec{d}}{\Vert d\Vert }\frac{{\vec{b}}^T\vec{d}}{\Vert d\Vert }{)}^T(\vec{b}-\frac{\vec{d}}{\Vert d\Vert }\frac{{\vec{b}}^T\vec{d}}{\Vert d\Vert })\ge 0\\ {\vec{b}}^T\vec{b}-\frac{({\vec{b}}^T\vec{d}{)}^2}{\Vert d{\Vert }^2}\ge 0\\ \Vert b{\Vert }^2\Vert d{\Vert }^2\ge (bd{)}^2\end{array}$$