Proof that $\|u+w\|^2 \geq \|u+w\|^2 - 2(\|v\|^2 - 1) + (\|v\|-1)^2 \geq 4 $

Question

Let $E$ be a Banach Space and $u,v, w \in E$ such that $\|u\| = 1 \leq \|v\|$ and $w = \frac{v}{\|v\|}$. I want to proof that $$\|u+w\|^2 \geq \|u+w\|^2 - 2(\|v\|^2 - 1) + (\|v\|-1)^2 \geq 4 $$

I got to proof only the first inequality:

since $\|v\|^2 - 1 = (\|v\| + 1)(\|v\|-1) \geq \|u+v\|\, (\|v\| - 1)$ and $\|w-v\| = \|v\| - 1$, $$ \|u+w\|^2 - 2(\|v\|^2 - 1) + (\|v\|-1)^2 \leq \left ( \|u+v\| - \|w-v\| \right )^2$$ as $\|u+v\| \leq \|u+w\| + \|w-v\|$, follows that $$ \left ( \|u+v\| - \|w-v\| \right )^2 \leq \|u+w\|^2 $$ and the first inequality holds.

However, I couldn't proof the second one. Help?

Answer 1

The second inequality does not hold with the stated conditions. For instance let $E=\mathbb C$, $u=1$, $v=-1$. Then $u+w=0$ and your second inequality is $0\geq 4$.