Proof that $ \|x+y\|^2 - \|x\|^2 \geq b(1 - 2^{-n})\|y\|^2 + 2^n( \|x+2^{-n}y\|^2 - \|x\|^2), \quad \forall x,y \in E, \forall n \in \mathbb N $ Suppose that $E$ is a Banach space over $\mathbb R$ satisfying the following inequality, for some $b > 0$
$$ \|x+y\|^2 + b\|x-y\|^2 \leq 2\|x\|^2 + 2 \|y\|^2, \quad \forall x,y \in E $$
I'm trying to proof that:
$$ \|x+y\|^2 - \|x\|^2 \geq b(1 - 2^{-n})\|y\|^2 + 2^n( \|x+2^{-n}y\|^2 - \|x\|^2), \quad \forall x,y \in E, \forall n \in \mathbb N \quad (I) $$
My attempt: An induction proof over $n$.
For $n = 1$, using the hypothesis replacing $x$ and $y$ for $(x+y)/2$ and $x/2$, respectively, we have:
$$ \|x + y/2\|^2 + b \|y/2\|^2 \leq 2\|(x+y)/2 \|^2 + 2\|x/2\|^2 = \frac{\|x+y\|^2}{2} + \frac{\|x\|^2}{2} $$
$$ \therefore \quad 2\|x+ 2^{-1}y\|^2 + b2^{-1}\|y\|^2 \leq \|x+y\|^2 + \|x\|^2 $$
$$ \therefore \quad \|x+y\|^2 - \|x\|^2 \geq  b2^{-1}\|y\|^2 +  2(\|x+ 2^{-1}y\|^2 - \|x\|^2) $$
which is the inequality $(I)$ for $n = 1$.
If I suppose that $(I)$ works for some $n > 1$, I have to proof that 
$$  \|x+y\|^2 - \|x\|^2 \geq b(1 - 2^{-n-1})\|y\|^2 + 2^{n+1}( \|x+2^{-n-1}y\|^2 - \|x\|^2), \quad \forall x,y \in E \quad (II) $$ 
However I don't know how to conclude that.
Thank you for any help!!

I think I might have solved the problem. Let's re-write (I) as
$$ \|x+y\|^2 - \|x\|^2 \geq b(\frac{2^n - 1}{2^n})\|y\|^2 + 2^n\|x+\frac{1}{2^n}y\|^2 - 2^n\|x\|^2 $$
$$ 2^n\|x+y\|^2 - 2^n\|x\|^2 \geq b(2^n - 1)\|y\|^2 + (2^n)^2\|x+\frac{1}{2^n}y\|^2 - (2^n)^2\|x\|^2 $$
$$  2^n\|x+y\|^2 - 2^n\|x\|^2 \geq b(2^n - 1)\|y\|^2 + \|2^n x+y\|^2 - \|2^n x\|^2 $$
Now, replacing $x$ for $2x$, we have
$$ 2^n\|2x+y\|^2 - 2^n\|2x\|^2 \geq b(2^n - 1)\|y\|^2 + \|2^{n+1} x+y\|^2 - \|2^{n+1} x\|^2 $$
$$ 2^{n+1} [2( \|x + y/2\|^2 -  \|x\|^2 ) ] \geq  b(2^n - 1)\|y\|^2 + (2^{n+1})^2\| x+\frac{1}{2^{n+1}}y\|^2 - (2^{n+1})^2\| x\|^2 $$
$$ 2( \|x + y/2\|^2 -  \|x\|^2 ) \geq b\frac{2^n - 1}{2^{n+1}}\|y\|^2 + 2^{n+1}\| x+\frac{1}{2^{n+1}}y\|^2 - 2^{n+1}\| x\|^2 $$
Summing $b\frac{1}{2}\|y\|^2$ in both sides:
$$ b\frac{1}{2}\|y\|^2 + 2( \|x + y/2\|^2 -  \|x\|^2 ) \geq b(\frac{1}{2}+\frac{2^n - 1}{2^{n+1}})\|y\|^2 + 2^{n+1}(\| x+\frac{1}{2^{n+1}}y\|^2 - \| x\|^2) $$
Applying the inequality for $n = 1$
$$ \|x+y\|^2 - \|x\|^2 \geq b(1-\frac{1}{2^{n+1}})\|y\|^2 + 2^{n+1}(\| x+\frac{1}{2^{n+1}}y\|^2 - \| x\|^2) $$
since $\frac{1}{2}+\frac{2^n - 1}{2^{n+1}} = \frac{1}{2} (1 + 1 - \frac{1}{2^n}) = (1 -\frac{1}{2^{n+1}})$
