Showing a Norm is not an Inner Product

[Punkyc7]

Show the taxicab norm is not an IP.

taxicab norm is v=(x$_{1}$...x$_{n}$)
then ||V||= |x$_{1}$|+...+|x$_{n}$|)

I was thinking about using the parallelogram law

but I would get this nasty thing(|x$_{1}$+w$_{1}$|+...+|x$_{n}$+w$_{n}$|)$^{2}$+(|x$_{1}$-w$_{1}$}+...+|x$_{n}$-w$_{n}$|$)^{2}$=2(|x$_{1}$|+...+|x$_{n}$|)$^{2}$+(|w$_{1}$|+...+|w$_{n}$|)$^{2}$Im not sure how to work with this. Am I going about this wrong?

Also there might be some typos with the absolute value signs the latex get messy

 

[dirk_mec1]

Hint: the norm doesn't satisfy linearity (it does satisfy the triangle inequality).

 

[clamtrox]

Do you mean that ||v+w||is not an inner product? This is easy to show! Just go through the definition of inner product -- one of the axioms is clearly not satisfied in this case.