I remember seeing a cool proof of the fact that norm is given by some inner product iff parallelogram law holds. It first showed that $||x+y||^2+||x-y||^2=2(||x||^2+||y||^2)$ implies $||x+y+z||^2=||x+y||^2+||y+z||^2+||z+x||^2-||x||^2-||y||^2-||z||^2$ (from this equality additivity of product given by polarization identity is trivial) (btw notice that this equality implies, and therefore is equivalent to parallelogram law; just substitute $z := -y$). I tried recreating this implication but I wasn't successful.
Also I was wondering whether inequality $||x+y||+||y+z||+||z+x|| \le ||x||+||y||+||z||+||x+y+z||$ is equivalent to parallelogram law (it's easy to see that it's true in inner product spaces).