I see that it fails to satisfy the triangle inequality by example but I don't see how to prove this is the case for all $0 < p < 1$. The definition I am using for $p$-norm is $$ \|A\|_p= \left(\sum_{k=1}^{n} |x_k|^p\right)^{1/p}.$$
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Yes, a short counter-example.
assume that $n=2$,take vectors $(1,0)$ and $(0,1)$.
you'll find it doesn't satisfy triangle inequality.

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