# $\left\lVert x \right\rVert_p$ = ($\sum_{i=1}^{n} |x_i|^p$)$^{1/p}$ and triangle inequality

How can I prove that for $$0 < p < 1$$ the defined function

$$\left\lVert x \right\rVert_p$$ = ($$\sum_{i=1}^{n} |x_i|^p$$)$$^{1/p}$$ for $$n \geq 2$$

does not fulfill the triangle inequality anymore?

I know that in one dimension, the norm is simply the absolute value and the triangle inequality holds. So here I could take two unit vectors.

For example, $$\vec{a}$$ = $$(0, 1, 0)$$ and $$\vec{b}$$ = $$(2/3, 2/3, 1/3)$$

But I don't really know how I can put the unit vectors into the sum above to show, that

$$\left\lVert a + b \right\rVert$$ $$\leq$$ $$\left\lVert a\right\rVert$$ + $$\left\lVert b \right\rVert$$ does not hold.

I have read this question yesterday night here, but it seems to be deleted, that's why I ask

You cannot add vectors in different dimensions. Your $$a$$ is in $$\mathbb R^{2}$$ and $$b$$ is in $$\mathbb R^{3}$$ so $$a+b$$ has no meaning. What you are asked to show is that if you fix $$n \geq 2$$ then we can have vectors $$a$$ and $$b$$ in $$\mathbb R^{3}$$ such that $$\|a+b\| >\|a\|+\|b\|$$. This is very easy: take the vectors $$(1,0,0,...,0)$$ and $$(0,1,0,...,0)$$ and verify that $$\|a+b\|=2^{1/p}$$, $$\|a\|+\|b\|=2$$. Since $$p<1$$ we have $$2^{1/p} >2$$.
Well, actually, this function is concave when you restrict your study to positive vectors (i.e., vectors with positive coordinates). If $$\nabla^2 f(x)$$ denotes the Hessian matrix of your function $$f(x)=\left(\sum_{i=1}^n x_i^p\right)^{1/p}$$ then, for any vector $$\vec{u}$$ you can show (after a lot of calculations) that $$\vec{u}^{T}\nabla^2 f(x)\vec{u}\leq 0,$$ hence the Hessian of $$-f$$ is positive semi-definite, so, by a known theorem, $$-f$$ convex, so $$f$$ is concave.