Linear algebra on norms

I am trying to show that the sup norm on $\mathbb{R}^2$ is not derived from the inner product on $\mathbb{R}^2$.

What I notice so far is that to suppose $\langle x,y\rangle$ is an inner product, not the dot product, and $|x| = \sqrt{\langle x,x\rangle}$. To compute $\langle x\pm y,x \pm y\rangle = \sqrt{(x \pm y)^2 + (x\pm y)^2}$. Also, what is special about saying $x=e_1$, and $y = e_2$?

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In your second line .Do you mean $|x|^{2}=<x,x>$? – Hassan Muhammad Nov 5 '11 at 19:53
Yes, but an inner product is a type of dot product – Buddy Holly Nov 5 '11 at 20:14

Hint: If a norm comes from an inner product, then it satisfies the parallelogram identity.

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Where, and how, does that identity fit in here? – Buddy Holly Nov 5 '11 at 21:05
@Buddy: Show the sup norm does not satisfy the parallelogram identity by selecting suitable (cleverly chosen) vectors and showing the identity is false. That will prove the sup norm does not derive from an inner product. – Arturo Magidin Nov 5 '11 at 22:01