Is the norm of a non-negative vector always smaller than the norm of the sum of two non-negative vectors? I am not a mathematician, so this question may be trivial, but I do not know whether it is correct or not. So any help is appreciated
Let $u$ and $v$ be two non-negative vectors, i.e., all elements in both vectors are non-negative.
Intuitively, I would say that the following inequality holds:$$\|u\| \leq \|u+v\|,$$
i.e., the norm over $u$ is smaller then the norm over the sum of $u$ and $v$. As the norm basically defines the 'length', the sum of non-negative vectors $u$ and $v$ should give a larger 'length'.
My question is, does this statement hold? And if so, what would be a proof?
Edit: I actually ment non-negative vectors instead of positive vectors. Thanks Rob Arthan for pointing this out.
 A: First, let me assume you are working with the usual Euclidean norm on the real $n$-dimensional vector space, $\Bbb{R}^n$. sLet $u = (u_1, \ldots, u_n)$ and $v = (v_1, \ldots, v_n)$ where the $u_i$ and $v_i$ are positive real numbers. Then for each $i$, $u_i < u_i+v_i$, so
$$
\|u\| = \sqrt{u_1^2 + \ldots u_n^2} < \sqrt{(u_1+v_1)^2 + \ldots (u_n+v_n)^2} = \|u + v\|
$$
If you want the analogous result with "non-negative" in place of "positive", just replace each $<$ above by $\le$.
You can ignore the rest of this answer if you are not interested in more general norms.
In general, a norm on $\Bbb{R}^n$ is completely determined by its unit disc, $D$, which can be an arbitrary convex body that is symmetric about the origin, i.e., $D = -D$. In $\Bbb{R}^2$ you can define a norm whose unit disc is the parallelogram with vertices at $(-1, 0)$, $(1, 1)$, $(1, 0)$ and $(-1, -1$). Under this norm $\|(0, 1/2)\| = 1 > \|(3/4, 3/4)\| = 3/4$.
A: It is not the case for all norms.  Consider the norm on $\mathbb R^2$ corresponding to the inner product
$\langle [x_1, x_2], [y_1, y_2] \rangle = x_1 y_1 -  (x_1 y_2 + x_2 y_1) + 2 x_2 y_2$
This is a norm because the matrix $\pmatrix{1 & -1\cr -1 & 2\cr}$ is 
positive definite.  
We have $$ \dfrac{\partial}{\partial y} \| [x,y] \|^2 = 
\dfrac{\partial}{\partial y} (x^2 - 2  x y + 2 y^2) = 
-2 x + 4 y$$
so e.g. taking $x = 1$ and $y = 0$, $\|[1,0] + [0,\epsilon]\| < \|[1,0]\|$ for small enough $\epsilon > 0$ (in fact for $0 < \epsilon < 1$).
EDIT: Ok, $[1,0]$ and $[0,\epsilon]$ don't have strictly positive entries, but by continuity the inequality will work for 
$[1,\delta]$ and $[\delta, \epsilon]$ if $\delta > 0$ is small.
A: The answer is no. Consider the norm $$\|(x,y)\| = |x| + |x-y|$$
on $\Bbb{R}^2$. Then
$$\|(1,1)\| = 1 < 2 = \|(1,0)\|.$$
