Parallelogram / Polarization Inequality 
Dear Math Experts,
as a stepping stone for another result I am interested in the following generalization of the parallelogram / polarization equality for Hilbert spaces. Denote by $H$ a Hilbert space over the real numbers $R$ and by $B = \{ u \in H : \Vert u \Vert \leq 1\}$ the closed unit ball in $H$.
The inner product is denoted by $\langle , \rangle$.
I am now after the following statement:
For any fixed vectors $x,y \in H$ we have
\begin{align}
-\Vert x - y \Vert^2 \leq 2\Vert x \Vert \langle u, x \rangle + 2\Vert y \Vert \langle u, y \rangle - \Vert x + y \Vert \langle u, x + y \rangle \leq \Vert x - y \Vert^2
\end{align}
for all $u \in B$.
For $u$ such that $\Vert u \Vert < 1$ equality holds if and only if $x = y$.

Any ideas related to my own proof strategy (see below) or otherwise are very welcome.
My own attempt so far:
I can deal with the cases $x=y$ and $x = 0 \neq y$ and I therefore assume $0 \neq x \neq y \neq 0$ in the following.
As a further observation we can regard the part in the middle as function of $u$ and rewrite in terms of the inner product as $f_{x,y}(u) = \langle u, (2\Vert x\Vert - \Vert x+y \Vert)x + (2\Vert y\Vert - \Vert x+y \Vert)y \rangle$.
Using Cauchy-Schwarz and $\Vert u \Vert \leq 1$ we then have 
\begin{align}
-L(x,y) \leq f_{x,y}(u) \leq L(x,y)
\end{align}
with $L(x,y) = \Vert (2\Vert x\Vert - \Vert x+y \Vert)x + (2\Vert y\Vert - \Vert x+y \Vert)y \Vert$.
The statement is then equivalent to $L(x,y) \leq \Vert x - y \Vert^2$.
Of course a result for general $H$ would be very interesting, but my own attempts have so far focused on the finite dimensional case $H = R^d$ with $\langle x,y \rangle = x^T y$ which is fine for my particular application.
We can now use that for any positive scalar $s$ and any rotation matrix $A$ we have $L(sAx,sAy) = s^2 L(x,y)$ as well as $\Vert sAx - sAy \Vert^2 = s^2 \Vert x - y\Vert^2$.
Assuming $0 \neq x \neq y \neq 0$ we then find a rotation matrix $A$ and the strictly positive scaling factor $s = 1 / \Vert x \Vert$ such that $sAx = e = (1,0,\ldots,0)$.
Wlog we then only consider $L(e,y) \leq \Vert e - y \Vert^2$ for any $y \in R^d$.
Therefore I need to show that $h(y) = \Vert e - y \Vert^2 - L(e,y) \geq 0$ at which point I am stuck.
I know that for $\alpha \geq 0$ we have $h(\alpha e) = 0$ which I believe is the lowest function value.
 A: Here is a proof of $L(x,y)\le |x-y|^2$ that is not very elegant (I use mathematica for the algebra) but at least works (so at least you know your claim is true!).
By the scale symmetry you observed we can wlog take $|x+y|=1$. Write $|x|=r\cos\theta, |y|=r\sin\theta$ for $0\le \theta\le \pi/2$ and $r\ge\frac{1}{\cos\theta+\sin\theta}$.
Since $|x+y|=1$ we have $r^2+2\langle x, y\rangle =1$ and so
$|x-y|^2=r^2-(1-r^2)=2r^2-1$.
And 
$L(x,y)^2=(1-2r\cos\theta)^2 r^2\cos^2\theta+(1-2r\cos\theta)(1-2r\sin\theta)(1-r^2)+(1-2r\sin\theta)^2r^2\sin^2\theta$.
Plugging into mathematica yields:
$L(x,y)^2-|x-y|^4=$ 
$-2(r\cos\theta+r\sin\theta-1)^2(2r^2\sin\theta\cos\theta+r\sin\theta+r\cos\theta)\le 0$.
A: Try taking an optimization approach. Take the Frechet derivatives of 
$$
x \mapsto L(x,y), \qquad y \mapsto L(x,y),
$$
and set them equal to $0$. You'll be able to take advantage of symmetry when doing so. This may put conditions on $x,y$ such that plugging them back into $L$ makes $L(x,y) \leq \|x-y\|^2$ apparent.
