Convex hull of sets defined by (in)equalities If you define the convex hull of a set $X$ as the set of all convex combinations of elements of $X$, it becomes difficult to decide if a given element $w$ belongs or not to $conv(X)$ (You have to discover whether  $w$  can be written as a convex combination of elements of X or not). But if you have a characterization of $conv(X)$ as a system of (in)equalities, it becomes easier.
Consider the sets:
$$A=\{(x,y,z)\in\mathbb{R}^3; y\geq1, z\geq1, y+z=3, x=3\} ,$$
$$B=\{(x,y,z)\in\mathbb{R}^3; x\geq1, y\geq1, x+y=3, z=3\}, $$
$$C=\{(x,y,z)\in\mathbb{R}^3; x\geq1, z\geq1, x+z=3, y=3\}. $$
These three sets are segments. Show that the convex hull  $\operatorname{conv}(A\cup B\cup C)$ is equal the set:
$$\begin{cases} x\geq1,y\geq 1,z\geq 1; \\ x+y\geq3, y+z\geq3, x+z\geq3;\\ x+y+z=6 \end{cases} $$
 A: Here's a boring way. Let $S$ denote the second set. It should be obvious that $A,B,C$ and $S$ are convex.
First notice that the extreme points of $A$ are $(3,1,2)$ and $(3,2,1)$, and that we can express $A = \mathbb{co} \{ (3,1,2), (3,2,1) \}$. It is easy to check that the extreme points are in $S$, hence $A \subset S$.
Similarly, the extreme points of $B$ and $C$ are the other combinations of $1,2,3$, and these can also be checked to be in $S$, hence $A \cup B \cup C \subset S$, hence by convexity $\mathbb{co} (A \cup B \cup C) \subset S$.
For the other inclusion, it helps if you sketch the set $S$, and see that it is an irregular convex hexagon in the $x+y+z=6$ plane. The extreme points will be seen to be all combinations of 1,2,3 as discussed above. I think it is easy to visualize by using the $x+y+z=6$ constraint to replace, for example, the constraint $x+y \geq 3$ by $z \leq 3$. Then $S$ can be seen to be the intersection of the plane $x+y+z=6$ and the constraints $x,y,z \in [1,3]$.
To complete the proof, one needs to verify that $S$ is the convex hull of the '$1,2,3$ combination points'. Let $(x,y,z) \in S$.
First suppose that $x=2$, then it is easy to verify that with $\lambda=\frac{y-1}{2} \geq 0$, then $(x,y,z) = \lambda (2,3,1)+(1-\lambda) (2,1,3)$.
Now suppose $x&lt2$. If you sketch $S$, it will be clear that $(x,y,z)$ lies in the convex hull of the points $p_1=(2,1,3)$, $p_2=(2,1,3)$, $p_3=(1,3,2)$ and $p_4=(1,2,3)$. It remains to find the coordinates.
A rather tedious calculation shows that with $\mu_1 = (x-1) \frac{(x+y-3)}{x}$, $\mu_2 = (x-1) (1-\frac{(x+y-3)}{x})$,
$\mu_3 = (2-x) \frac{(x+y-3)}{x}$, $\mu_4 = (2-x) (1-\frac{(x+y-3)}{x})$, we have $(x,y,z) = \sum_{i=1}^4 \mu_i p_i$, with
$\mu_i \geq 0$ and $\sum_{i=1}^4 \mu_i =1$. Hence $(x,y,z) \in \mathbb{co} \{ p_i \}_{i=1}^4$.
A similarly boring calculation shows a similar result for $x>2$. Hence $S$ is the convex hull of the '$1,2,3$ combination points'.
