Is the norm of a non-negative vector always smaller than the norm of the sum of two non-negative vectors?

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 $\|(0, 1/2)\| = 1 > \|(3/4, 3/4)\| = 3/4$.

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.