Let's consider inner product space with vectors $x, y, z$ which satisfies:
$$\|x+y+z\|^2 = 14$$
$$\|x+y-z\|^2 = 2$$
$$\|x-y+z\|^2 = 6$$
$$\|x-y-z\|^2 = 10$$
I want to prove that $x$ is perpendicular to $y$.
My work so far
In other words we want to prove that $\langle x, y \rangle = 0$.
My first idea was to use Cauchy Schwarz inequality $ | \langle x , y \rangle |\le \|x\| \|y\|$ and to show that our conditions force that $\|x\|\cdot\|y\| = 0$. However I didn't manage to do anything sensible. Also I tried to somehow prove that under this conditions our norm has bo inducted by inner product - in other words we have that: $$2(\|x\|^2 + \|y\|^2) = \|x+y\|^2 + \|x-y\|^2$$
but also I didn't end up with something rational. Could you please give me a hand, what's the correct approach to this problem?