= Solution
Expand using the bilinearity and symmetry of the real <inner product>:
$$
\|x+y\|^2=\langle x,x\rangle+2\langle x,y\rangle+\langle y,y\rangle,\qquad
\|x-y\|^2=\langle x,x\rangle-2\langle x,y\rangle+\langle y,y\rangle.
$$
Addition proves the <parallelogram law>,
$$
\boxed{\|x+y\|^2+\|x-y\|^2=2\|x\|^2+2\|y\|^2.}
$$
For the maximum <norm> on $\mathbb R^2$, take $x=(1,0)$ and $y=(0,1)$. All four vectors $x,y,x+y,x-y$ have maximum <norm> one. The left side of the <parallelogram law> is then $2$, whereas the right side is $4$. \b[The maximum norm cannot come from an inner product.]
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