Example of norm on $\mathbb{R}^2$ that's NOT absolutely monotonic. We call a norm $\|\cdot\|$ on $\mathbb{R}^n$ absolutely monotonic if 
$$
|a_i| \leq |b_i|, i=1,\cdots,n \implies \|a\| \leq \|b\|.
$$
What's an example of norm on $\mathbb{R}^2$ that's not absolutely monotonic?

This is an exercise left to the reader -- so presumably not too difficult. But it's giving me some trouble.
The usual suspects that come to mind, i.e. the $\ell^p$ norms, $p \in [1,\infty]$, are all absolutely monotonic.
 A: Consider a sheer $A(x,y):=(x,y-x)$ and then the norm $\|(x,y)\|:=\|A(x,y)\|_1$. Then $\|(1,0)\|=2$ but $\|(1,1)\|=1$.
Interestingly though, for every norm $\|\cdot\|:\mathbb{R}^2\to[0,\infty)$ there exists a linear map $A:\mathbb{R}^2\to\mathbb{R}^2$ such that $\|\cdot\|\circ A$ is an absolutely monotonic norm.
To prove this, note that a norm is absolutely monotonic if (and only if, may I add) the smallest axis-aligned rectangle containing the unit disk touches the unit circle at the intersections with the $x$- and $y$-axis. This is a consequence of the triangle inequality. So if we find vectors $v$ and $u$ on the unit sphere such that the unit disk is contained in the parallellogram $[-1,1]v\times[-1,1]u$, then the unique linear map that sends $v$ to $(1,0)$ and $u$ to $(0,1)$ suffices.
In order to find such $v$ and $u$, we will use the intermediate value theorem. For any angle $\theta$ let $v(\theta)$ be the unique vector on the unit sphere with angle $\theta$ with respect to the origin. The uniqueness follows from the scaling property and the non-degeneracy of norms, and the continuity follows from the triangle inequality. Then let $u(\theta)$ be the vector on the unit sphere furthest to the left of the line generated by $v(\theta)$. A little bit of a problem here is that $u(\theta)$ is not always uniquely defined, and $\theta\mapsto u(\theta)$ is only continuous at points where it is uniquely defined. However, for now just pretend that it is a well-defined continuous function, and I will come back to this later.
We now find that $\mathbb{R}v\times[-1,1]u$ contains the unit disk. Most importantly the intersection of the unit disk with $(1,\infty)v\times[-1,1]u$ is fully contained in either $(1,\infty)v\times[-1,0]u$ or $(1,\infty)v\times[0,1]u$, due to the triangle inequality. If for some $\theta$ this intersection lies in one of the two, then if we move $\theta$ such that the new $v(\theta)$ is where the old $u(\theta)$ was, the intersection will lie in the other one. For some $\theta$ in between, we must therefore have that the intersection lies in neither. Therefore the unit disk will be contained in $[-1,1]v\times[-1,1]u$.
Finally, to come back to the little problem. For such problematic $\theta$ you can imagine fixing $v(\theta)$ while moving $u(\theta)$ continuously along the set of all possible values.
This is a little bit heuristic, but I hope it is convincing anyways.
A: I would like to give an answer similar to Smiley-Craft but from a different point of view:
The standard norm on $\mathbb{R}^2$ is given by $\|(a,b)\|= \sqrt {a^2+b^2}$. We can define a similar norm using a different basis.
Given $u,v$ a basis of $\mathbb{R}^2$. We can define the norm of $w$ by $$\|w\| = \sqrt{a^2+b^2}$$ where $w=au+bv$.
It is an easy exercise to show that this is indeed a norm (same proof as in the standard norm case).
Since we can change coordinates it will be really easy to define a non-monotonic norm:
For instance choose $u=(\frac{1}{2},0)$ and $v=(1,1)$ then $\|(1,0)\| = 2$ while $\|(1,1)\| = 1$.
Side note: It is an interesting question whether every norm is absolutely monotonic with respect to some basis. I would guess the answer is yes but I don't know how to approach this question yet.
A: Here's a way to come up with an answer using as little geometric intuition as possible.
The standard norm on $\mathbb{R}^2$ can be written as:
$$ \left\Vert \begin{pmatrix} a \\ b \end{pmatrix} \right\Vert^{\;2} \mapsto \begin{pmatrix}a \\ b\end{pmatrix}^T \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} \begin{pmatrix} a\\ b \end{pmatrix} \tag{1} $$
or
$$ \left\Vert \begin{pmatrix} a \\ b \end{pmatrix} \right\Vert^{\;2} \mapsto a^2 + b^2 \tag{2} $$
Which makes it immediately clear that, for arbitrarily chosen $(a_1, b_1)$ and $(a_2, b_2)$ .
$$ \frac{a_1^2 \le a_2^2 \;\;\;\;\text{and}\;\;\;\; b_1^2 \le b_2^2}{a_1^2 + b_1^2 \le a_2^2 + b_2^2} \tag{3} $$
If we stare at (1) and (2), it suggests that by adding a cross term we can "penalize" negative components, so let's pick a simple one, let's call the new norm $\nu$.
$$ \left\Vert \begin{pmatrix} a \\ b \end{pmatrix} \right\Vert^{\;2}_{\,\nu} \mapsto \begin{pmatrix}a \\ b\end{pmatrix}^T \begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix} \begin{pmatrix} a\\ b \end{pmatrix} \tag{4} $$
$$ \left\Vert \begin{pmatrix} a \\ b \end{pmatrix} \right\Vert^{\;2}_{\,\nu} \mapsto a^2 + ab + b^2 \tag{5} $$
Let's pick $(a_1, b_1) = (2, 2)$ and $(a_2, b_2) = (-3, 3)$ .
$$  \left\Vert \begin{pmatrix} 2 \\ 2 \end{pmatrix} \right\Vert^{\;2}_{\,\nu} = 4 + 4 + 4 = 12 \tag{6} $$ 
$$  \left\Vert \begin{pmatrix} -3 \\ 3 \end{pmatrix} \right\Vert^{\;2}_{\,\nu} = 9 - 9 + 9 = 9 \tag{7} $$ 
So, $(2, 2)$ and $(-3, 3)$ indeed demonstrates that $\nu$ is not absolutely monotonic.
All that's left is to verify that it's an actual norm.
The norm axioms are:
$$ \nu\,(\alpha v) = |\alpha| \,\nu\,(v) \tag{8} $$
$$ \nu\,(u+v) \le \nu\,(u) + \nu\,(v) \tag{9} $$
$$ \nu\,(v) = 0 \implies v = \vec{0} \tag{10} $$
(8) falls out of the fact that matrix multiplication is linear. (10) falls out of the invertibility of $\left[\begin{smallmatrix}1& 1\\0& 1\end{smallmatrix}\right]$
I think the most straightforward way to prove (9) is to note that the matrix  $\left[\begin{smallmatrix}1& 1\\0& 1\end{smallmatrix}\right]$ is upper triangular and hence its eigenvalues with multiplicity are the multiset $\{1, 1\}$ . Since all of its eigenvalues are positive, the matrix is positive definite and therefore the quadratic form $\nu\,(u) = u^T \left[\begin{smallmatrix}1 & 1 \\0 & 1\end{smallmatrix}\right] u$ is a norm.
