# Trouble understanding proof of the Cauchy-Schwarz inequality

Hope you're doing well.

Well I'm stuggling to understand the proof of Cauchy-Schwarz inequality, especially with the following equation: $$\frac{1}{\|\mathbf{v}\|^{2}}\|\| \mathbf{v}\left\|^{2} \mathbf{u}-\langle\mathbf{u}, \mathbf{v}\rangle \mathbf{v}\right\|^{2}=\|\mathbf{u}\|^{2}\|\mathbf{v}\|^{2}-|\langle\mathbf{u}, \mathbf{v}\rangle|^{2}$$

As it says in wikipedia: if you simplify the right side of the equation you'll get the left side easily. I tried and it doesn't work for me.

• note that $\Vert x\Vert^2 = \langle x,x\rangle$ Jan 24, 2021 at 22:13
Note that\begin{align}\bigl\|\| \mathbf{v}\|^{2} \mathbf{u}-\langle\mathbf{u}, \mathbf{v}\rangle \mathbf{v}\bigr\|^2&=\|\mathbf v\|^4\|\mathbf u\|^2-2\|\mathbf v\|^2\langle\mathbf u,\mathbf v\rangle^2+\bigl|\langle\mathbf u,\mathbf v\rangle\bigr|^2\|\mathbf v\|^2\\&=\|\mathbf v\|^2\bigl(\|\mathbf v\|^2\|\mathbf u\|^2-\bigl|\langle\mathbf u,\mathbf v\rangle\bigr|^2\bigr).\end{align}Now, divide both sides by $$\|\mathbf v\|^2$$ and you will get your equality.
• Is the following statement true (because the inner product doesn't have to be a dot product): $\langle\mathbf{u}, \mathbf{u}\rangle=\|\mathbf{u}\|^{2}$ Jan 25, 2021 at 8:27