Proof that $\frac{b}{1+b} \|u-v\| > \|v\| - 1$ and $\|u-w\| \geq (1+b)^{-1} \|u-v\|$

Question

Let $E$ is a Banach Space, let $u,v \in E$ and $w = \frac{v}{\|v\|}$ s.t. $\|u\|= 1 \leq \|v\|$. Taking $c = b/(1+b)$ suppose that $$\|u+v\|^2 + c^2 \|u-v\|^2 > 2 + 2\|v\|^2 \quad \quad (1) $$

I'm trying to proof that $\frac{b}{1+b} \|u-v\| > \|v\| - 1$ and $\|u-w\| \geq (1+b)^{-1} \|u-v\|$.

My attempt:

The first inequality follows from triangle inequality, indeed $$ c^2 \|u-v\|^2 > 2 + 2\|v\|^2 - \|u+v\|^2 \geq 2+ 2\|v\|^2 - \|u\|^2 - \|v\|^2 - 2\|v\| $$ since $\|u\|=1$, follows $$ c^2 \|u-v\|^2 > 1 + \|v\|^2 - 2\|v\| = (\|v\| - 1)^2$$ hence, $c \|u-v\| >\|v\| - 1$, where $c = \frac{b}{1+b}$.

The second inequality I can't see why it holds only using $(1)$.

Help?

Answer 1

I will assume that the intent is that $b>0$ and that $c>0$. With that I mind, I would think that in the second inequality, it should be $1+b$ instead of $1-b$ (otherwise, for instance, the natural case $c=1/2$ leads to an impossible second inequality).

We have, using the triangle inequality and then the first inequality, \begin{align} \|u-v\|&\leq \|u-w\|+\|w-v\|=\|u-w\|+(\|v\|-1)\\ \ \\ &<\|u-w\|+\frac b{1+b}\,\|u-v\|. \end{align} Then $$ \|u-w\|>\left(1-\frac b{1+b}\right)\,\|u-v\|=(1+b)^{-1}\,\|u-v\|. $$