$\frac{2}{\lVert x \rVert^2} \langle x,u \rangle \langle x,v\rangle - \langle u, v\rangle \leq \lVert u \rVert \lVert v \rVert$?

Question

In $\mathbb R^n$, $\frac{2}{\lVert x \rVert^2} \langle x,u \rangle \langle x,v\rangle - \langle u, v\rangle \leq \lVert u \rVert \lVert v \rVert$?

The book says the above inequality can be obtained by using Cauchy-Schwarz inequality and the bound $$\langle R(u), v \rangle \leq \lVert R(u) \rVert \lVert v \rVert \leq \lVert u \rVert \lVert v \rVert, $$ where $R(u)$ is the reflection point of the vector $u$. I don't see how to get the desired inequality from the hint.

Answer 1

Let $\hat x = \frac{x}{|x|}$. Then $$\begin{align} 2\langle \hat x, u\rangle\langle \hat x, v\rangle - \langle u, v\rangle &= \langle 2\hat x\langle \hat x,u\rangle-u,v\rangle \\ &= \langle R(u),v\rangle \end{align}$$ because as N8tron said, $R(u)=2\hat x\langle \hat x,u\rangle-u$.

For completeness: $$\begin{align} |R(u)|^2 &= |\hat x\langle \hat x,u\rangle+(\hat x\langle \hat x,u\rangle-u)|^2\\ &=\langle \hat x,u\rangle^2+|\hat x\langle \hat x,u\rangle-u|^2 \tag{Pythagoras's thm}\\ &=\langle \hat x,u\rangle^2+|u-\hat x\langle \hat x,u\rangle|^2\\ &= |\langle \hat x,u\rangle\hat x+u-\hat x\langle \hat x,u\rangle|^2 \tag{Pythagoras's thm}\\ &=|u|^2 \end{align}$$