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