Finding the set of parameters for which the inequality holds for all $x,y$ I encountered the following question about inequalities which I am curious how to solve.
The simplest case is to consider the inequality
$$|x|+|y|+|x+y|+ax+by\geq 0$$
where $x,y,a,b\in\mathbb{R}$.
The question is which values of parameters $a,b\in\mathbb{R}$ would guarantee that the inequality holds for all $x,y\in\mathbb{R}$.
I tried playing with such inequalities and was surprised that this question is harder than I anticipated.
I first found that necessarily $a\in [-2,2]$:
for $x=1$ and $y=0$ we get $2+a\geq 0$ so $a\geq -2$ and for $x=-1$ and $y=0$ we get $2-a\geq 0$ so $a\leq 2$.
Similarly, $b\in [-2,2]$ must hold.
Furthermore, I did find out that if $a=b\in [-2,2]$ then the inequality certainly holds for all $x,y\in\mathbb{R}$, because
$$|x|+|y|+|x+y|+ax+by=(|x|+ax/2)+(|y|+ay/2)+(|x+y|+a(x+y)/2)$$
and each element in the parentheses is non-negative because $a/2\in [-1,1]$ and so $$|z|+az/2=\begin{cases} (1+a/2)z & z\geq 0\\ (-1+a/2)z & z<0\end{cases}\geq 0$$
However, I am sure that there are also other parameters $a,b\in [-2,2]$ for which the inequality holds. For instance $a=2$ and $b=1$ turn out to be a valid solution for all $x,y\in\mathbb{R}$ if you consider the various different cases.
My question is how would I go about finding all such $a,b$?
More generally, is there some theory about such inequalities? Any good reference (paper/book) that is recommend to read in order to better grasp these inequalities?
I am also interested in more general inequalities, with more parameters and variables, for instance
$|x|+|y|+|z|+|x+y|+|x+z|+ax+by+cz\geq 0$
(with the task of finding $a,b,c$ so that it holds for all $x,y,z$).
I am interested to which extent there is a theory of how to solve these questions generally.
 A: Suppose we have $\def\sgn{\operatorname{sgn}}$
$$|x|+|y|+|x+y|+ax+by \geqslant 0 \tag 1$$
for all $x$ and $y$.  Then this must apply in particular for $y=0$:
$$2|x|+ax = |x|(2+a\sgn x) \geqslant 0 
\quad\implies\quad 2+a\sgn x \geqslant 0\tag 2$$
where $\sgn x$ denotes the sign of $x$.  This implies
$|a|\leqslant 2$ and also $|b|\leqslant2$ due to symmetry. When used with $x=-y$, one gets $|a-b|\leqslant 2$.
So more generally, let's try $y=wx$, so that $(1)$ becomes
$$\begin{align}
& |x|+|wx|+|x+wx|+ax+bwx \\
& \quad = |x|(1+|w|+|1+w| + (a+bw)\sgn x) \geqslant 0 \tag 3
\end{align}$$
Divide $(3)$ by $|x|$ to get
$$(a+bw)\sgn(-x) \leqslant 1+|w|+|1+w| \tag 4$$
resp.
$$|a+bw| \leqslant 1+|w|+|1+w| \qquad\text{for all }w\tag 5$$
Thus in the plane, the line $a+wb$ is wedged between $\pm(1+|w|+|1+w|)$.  The wedge is shown in the graphic in red+blue.  However, this wedging is too restrict because it does not take into account that both line and wedge depend on $w$, i.e. the wedge is a bit like a moving target. Lines that do not cross a wedge still represent valid solutions, though.

Now $(5)$ is simpler than $(1)$ in the sense that we have only one value, $w$, in the absulute values on the right side, so that when cases are considered, we are left with just 3 cases instead of the 8 cases of $(1)$. So let's try and split cases:
$$|a+wb|\leqslant\begin{cases}
2+2w, & w\geqslant 0 \\
2, & -1\leqslant w \leqslant 0 \\
-2w, & w \leqslant -1 \\
\end{cases}$$
The constraints from above are recovered for $w\in\{0,-1,\infty\}$.
Notice that due to symmetry, this must still hold true if $a$ and $b$ are exchanged.  Whilst this is somewhat simpler than $(1)$, it's not a satisfying solution and I am stuck here, but maybe some ideas are helpful.

Playing around with Desmos, it appears that the set you are looking for is the hexagon described by
$$\{a,b\in[-2,2] \ :\ |a-b|\leqslant 2\}$$
https://www.desmos.com/calculator/snisn01tvk
The parameter $w$ is named "g" in Desmos and can be changed with a slider.  The plane is the $a$-$b$-plane.  The set of solution is the intersection over all "g" resp. $w$.
A: Letting $x = 1, y = 0$, we have
$2 + a \ge 0$.
Letting $x = -1, y = 0$, we have
$2 - a \ge 0$.
Letting $x = 0, y = 1$, we have
$2 + b \ge 0$.
Letting $x = -1, y = 1$, we have
$2 - a + b \ge 0$.
Letting $x = 0, y = -1$, we have
$2 - b \ge 0$
Letting $x = 1, y = -1$, we have
$2 + a - b \ge 0$.
Thus, we have $a, b \in [-2, 2]$ and $|a - b| \le 2$.
$\phantom{2}$
On the other hand, suppose that
$a, b \in [-2, 2]$ and $|a - b| \le 2$.
Let us prove that
$$|x| + |y| + |x + y| + ax + by \ge 0,
\, \forall x, y \in \mathbb{R}.$$
We split into three cases:
Case 1: If $y = 0$, we have
$$\mathrm{LHS} = 2|x| + ax \ge 2|x| - |a|\cdot |x| = (2 - |a|)|x| \ge 0.$$
Case 2: If $y > 0$, since the inequality is homogeneous, WLOG, assume that $y = 1$.
It suffices to prove that
$$|x| + 1 + |x + 1| + ax + b \ge 0, \, \forall x\in \mathbb{R}.$$
(1) If $x \le -1$, we have
$$\mathrm{LHS} = - (2 - a)x + b \ge (2 - a) + b \ge 0.$$
(2) If $-1 < x \le 0$, we have
$$\mathrm{LHS} = 2 + ax + b.$$
If $a \le 0$, we have $2 + ax + b \ge 0$.
If $a > 0$, we have $2 + ax + b \ge 2 - a + b \ge 0$.
(3) If $x > 0$, we have
$$\mathrm{LHS} = (2 + a)x + 2 + b \ge 0.$$
Case 3: If $y < 0$, since the inequality is homogeneous, WLOG, assume that $y = -1$.
It suffices to prove that
$$|x| + 1 + |x - 1| + ax - b \ge 0, \, \forall x\in \mathbb{R}.$$
(1) If $x \le 0$, we have
$$\mathrm{LHS}
= - (2 - a)x + 2 - b \ge 0.$$
(2) If $0 < x \le 1$, we have
$$\mathrm{LHS} = 2 + ax - b.$$
If $a \ge 0$, we have $2 + ax - b \ge 0$.
If $a < 0$, we have $2 + ax - b \ge 2 + a - b \ge 0$.
(3) If $x > 1$, we have
$$\mathrm{LHS}
= (2 + a)x - b \ge 2 + a - b \ge 0.$$
We are done.
