Let $V$ a Banach Space over $\mathbb R$ and $0 < b < 1$. Assuma that, for all $x,y \in V$ with $\|x\|=\|y\| = 1$, we have $$ \|x+y\|^2 + b^2\|x-y\|^2 \leq 4 \quad (I) $$
I'm trying to proof that for $c = b/(1+b)$ $$ \|x+y\|^2 + c^2\|x-y\|^2 \leq 2\|x\|^2 + 2\|y\|^2, \quad \forall x,y \in V \quad (II)$$
My attempt: Suppose, by contradiction, that $(II)$ is false. Then, w.l.o.g., $x,y \in V$ s.t. $1 = \|u\| \leq \|v\|$ satisfying $$ \|u+v\|^2 + c^2\|u-v\|^2 > 2\|u\|^2 + 2\|v\|^2 = 2 + 2\|v\|^2\quad (1)$$ Then $$ c^2\|u-v\|^2 > 2 + 2\|v\|^2 - \|u+v\|^2 \geq 2 + 2\|v\|^2 - (1 + \|v\|)^2 $$ Since $\|u+v\| \leq \|u\| + \|v\| = 1 + \|v\|$. And, hence $$ c^2\|u-v\|^2 > 2 + 2\|v\|^2 - (1 + \|v\|^2 + 2\|v\|) = \|v\|^2 - 2 \|v\| + 1 = (\|v\| - 1)^2$$ $$ \therefore \quad c\|u-v\| > \|v\| - 1 \quad (2) $$ Consider $w = \frac{v}{\|v\|}$. The next step in the proof should be proof that $$\|u+w\|^2 + b^2 \|u-w\|^2 \geq \|u+v\|^2 - 2(\|v\|^2 - 1) + (\|v\| - 1)^2 + c^2 \|u-v\|^2 \quad (\ast) $$ And this would complete the proof since $$ 4 \geq \|u+w\|^2 + b^2 \|u-w\|^2 \geq \|u+v\|^2 - 2(\|v\|^2 - 1) + (\|v\| -1)^2 + c^2 \|u-v\|^2 > 4 + (\|v\| - 1)^2 $$ Since $\|u+v\|^2 - 2(\|v\|^2 - 1) + (\|v\| -1)^2 \geq 4$ and $(2)$, and we obtain a contradiction.
I would like a help to see why the inequality $(\ast)$ is true. Thank you for the help.