Parallelogram / Polarization Inequality

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:

$L(x,y)^2-|x-y|^4=$

$-2(r\cos\theta+r\sin\theta-1)^2(2r^2\sin\theta\cos\theta+r\sin\theta+r\cos\theta)\le 0$.