Let $E$ be a normed space and let a vector $u\in E$ and a vector $v\in E$ with $||v||=1$ that satisy $||u+v||\geq 2-\varepsilon $ for a $\varepsilon >0$ be given. Show that for all $\alpha ,\beta >0$ the following holds:
$$||\alpha u+\beta v||\geq (1-\varepsilon )(\alpha +\beta )$$.
I have been stuck with this question for days and this is how far I've come:
$||\alpha u+\beta v||\leq ||\alpha u||+||\beta v||\\ =|\alpha |||u||+|\beta |||v||\\ =\alpha ||u||+\beta ||v||\\ =\alpha ||u||+\beta \\ \Rightarrow ||\alpha u+\beta v||\leq \alpha ||u||+\beta$
$||u+v||\leq ||u||+||v||=||u||+1\\ \Rightarrow 2-\varepsilon\leq ||u+v||\leq ||u||+1\\ \Rightarrow 2-\varepsilon \leq ||u||+1\\ \Rightarrow 1-\varepsilon \leq ||u||\\ \Rightarrow (\alpha +\beta )(1-\varepsilon )\leq ||u||(\alpha +\beta )$
Which seems to be very close, but it's not yet there and I can't get any longer.
A combination of the above (assuming that the inequality I'm supposed to show holds true) gives:
$\alpha ||u||+\beta ||u||\leq \alpha ||u||+\beta \\ \beta ||u||\leq \beta \\ ||u||\leq 1$
The definition of a norm in combination with what's given in the question should also lead to $0<\varepsilon \leq 2$.
But that's it, I just can't prove it. I have managed to show that it holds true for $\varepsilon =2$, but the assumption should be for a given $\varepsilon >0$, not one that I assume myself.