Finding the set of parameters for which the inequality holds for all $x,y$

Question

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.

Answer 1

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.

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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. $w$.

Answer 2

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.