Let $(V,\langle,\rangle)$ be a euclidean vector space and $u, v \in V$ two vectors. With $||x||=\sqrt{\langle x,x\rangle}$ the parallelogram law $||u+v||^{2} + ||u-v||^{2} = 2||u||^{2} + 2||w||^{2}$ is valid.
(a)Prove the parallelogram law.
(b)Let $V=\mathbb{R}^{2}$ with the canonical scalar product. Provide a geometric proof of the parallelogram law (Pythagoras).
(c)Show that the norm $||x|| = \displaystyle\sum\limits_{i=1}^n|x_i|$ on $\mathbb{R}^{2}$ does not come from a scalar product.
I guess (a) was just a matter of expanding the terms and using bilinearity/symmetry:
\begin{align*} ||u+v||^{2}+||u-v||^{2} &= \langle u+v,u+v\rangle+\langle u-v,u-v\rangle\\ & = \langle u,u+v\rangle+\langle v,u+v\rangle+\langle u,u-v\rangle-\langle v,u-v\rangle\\ &= \langle u+v,u\rangle+\langle u+v,v\rangle+\langle u-v,u\rangle-\langle u-v,v\rangle \\ &= \langle u,u\rangle+\langle v,u\rangle+\langle u,v\rangle+\langle v,v\rangle+\langle u,u\rangle-\langle v,u\rangle-\langle u,v\rangle+\langle v,v\rangle \\ &= 2\langle u,u\rangle+2\langle v,v \rangle\\ &= 2||u||^{2}+2||v||^{2}. \end{align*}
For(c): is my goal to show that the norm is either not symmetric, bilinear, or positive definite? I am not sure how to show this since it does not say what to do with $y$, (my understanding is that a scalar product is a numerically valued function of ordered pairs of vectors $(x,y)$ which satisfy those three conditions). I am guessing however, that it is not enough to just write that this norm is not a scalar product since it doesn't tell me what to do with $y$, right?
With (b): all I know is that the canonical scalar product is $\langle x,y\rangle = \displaystyle\sum\limits_{i=1}^nx_iy_i$, but what exactly do they mean by a "geometric" proof?
Thank you in advance for any help!