Conditions on the halfspace $ax+by+cz>d$ What conditions do we have to set on the halfspace $\{(x,y,z) \in \mathbb{R}^3 \vert ax+by+cz>d \}$ ($a>0$) for that it contains the half space $ \{(x,y,z) \in \mathbb{R}^3  \vert x \geq 0 \}$?
Since $(0,0,0)$ is then contained in it, we must have that $d < 0$, thus $a^{-1}d < 0$. What can I say about $b$ and $c$?
 A: Initial remarks:

*

*The "contains" condition can be translated into the equivalent one:

$$\{ x | x \ge 0\} \cap \{ x | ax+by+cz-d<0\} = \emptyset \tag{1}$$


*If two distinct planes with equations $ux+vy+wz=d$ and  $u'x+v'y+w'z=d'$ are not parallel, the intersection of halfspaces $ux+vy+wz>d$ and  $u'x+v'y+w'z<d'$ (made of two "wedges") is non void (a classical convexity result).

As a consequence of Remark n°2, the only possibility to have relationship (1) is to have nonintersecting, planes as boundaries, i.e., parallel planes ; a parallel plane to $x=0$ is a plane with equation $x=d$.
Therefore, only halfspaces defined by an inequation having the form $x > d$ can contain halfspace $x>0$. Moreover, this $d$ must be negative.
In particular, $b$ and $c$ must be zero.
Remark: A distinction should be established between closed and open halfspaces. I should have considered this distinction, but I did not want to blur the message.
A: I am answering assuming you meant how to cover the set $A=\{(x,y,z)\in \mathbb{R}^3, x\ge 0\}$ by a family of planes. The set $\{(x,y,z), ax+by+cz>d\}$  covers $A$ if and only if $b=c=0$ and $d<0$. Any inclined family of planes (when $b$ and $ c$ are non zero) for any fixed $d$ cannot contain all positive abscisses points, for $d$ fixed you may take  large values of $y$ or $z$ (depending on $a,b,c$ signs) and small values of positive $x$ to prove that.
