I have found a following statement. Let $E$ be a real normed linear space.
Prove that $E$ is an inner product space if and only if for any $x,y \in\mathbb E$ there is some $\alpha \in \mathbb R$ that $\|(1-\lambda \alpha)x-\lambda y\|=\|(1+\lambda \alpha)x+\lambda y\|$ for all $\lambda \in \mathbb R$.
The first part is easy: I suppose that $E$ is an inner product space, then I get from the squared equality
$$(1-\lambda\alpha)^2\langle x,x\rangle+2\lambda(1-\lambda\alpha)\langle x,y\rangle=(1+\lambda\alpha)^2\langle x,x\rangle+2\lambda(1+\lambda\alpha)\langle x,y\rangle$$
which leads to $$\alpha=-\frac{\langle x,y\rangle}{\langle x,x\rangle}.$$
But now suppose that I have that for any $x,y \in\mathbb E$ there is some $\alpha \in \mathbb R$ that $\|(1-\lambda \alpha)x-\lambda y\|=\|(1+\lambda \alpha)x+\lambda y\|$ for all $\lambda \in \mathbb R$.
I would like to use the parallelogramm identity: $\|x+y\|^2+\|x-y\|^2=2\|x\|^2+2\|y\|^2$, but I cannot see how to do it.