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

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.

• What's $(x,v)$? Jun 5 '18 at 15:26
• @ogogmad Thanks! Fixed. Jun 5 '18 at 15:29
• Reflection of $v$ across a hyperplane orthogonal to $x$ through the origin would be $$v-2\frac{\left<v,x\right>}{\left<x,x\right>}x$$ Jun 5 '18 at 15:32

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}