Norm and InnerProduct Inequality

Question

How can I show that this is true:

Let $u,v \in \mathbb{R}^n$: \begin{align} \frac{\|u\|}{\|v\|} \leq \frac{(u,u-v)}{(v,u-v)}, \quad \hbox{if} \quad (v,u-v) > 0 \end{align} Where $\|\cdot\|$ is the induced norm to the inner product $(\cdot \ ,\cdot)$ At the moment I have no clue how to obtain this result. It is somehow connected to a geometric interpretation of a triangle?

Thanks a lot!

Answer 1

$$\begin{align*}(u,u-v)=||u||^2-(u,v)\\ (v,u-v)=||v||^2-(u,v)\end{align*}\implies$$

$$ \left(\frac{(u,u-v)}{(v,u-v)}\right)^2=\frac{||u||^4-2(u,v)||u||^2+(u,v)^2}{||v||^4-2(u,v)||v||^2+(u,v)^2}\ge\frac{||u||^2}{||v||^2}\iff$$

$$||u||^4||v||^2-\color{red}{2(u,v)||u||^2||v||^2}+(u,v)^2||v||^2\ge||u||^2||v||^4-\color{red}{2(u,v)||u||^2||v||^2}+(u,v)^2||u||^2$$

$$\iff||u||^2||v||^2\left(||u||^2-||v||^2\right)\ge(u,v)^2\left(||u||^2-||v||^2\right)$$

The inequality is clear if $\;||u||=||v||\;$ , otherwise we get from the above that our inequality is true iff

$$||u||^2||v||^2\ge(u,v)^2$$

and now C-S kicks in ...

Answer 2

\begin{align} \frac{\|x\|}{\|y\|} \leq \frac{(x,x-y)}{(y,x-y)} = \frac{\|x\|^2- \|x\|\|y\|\cos \varphi}{\|x\|\|y\|\cos \varphi - \|y\|^2 } = \frac{\frac{\|x\|}{\|y\|}- \cos \varphi}{\cos \varphi - \frac{\|y\|}{\|x\|}} \end{align} The minimum of this function is attained at $\cos \varphi = 1$ with the value $\frac{\|x\|}{\|y\|}$. Therefore this value is a lower bound!