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.

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.
