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 $x<2$. 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'.
There is a theorem in van de Vel's Theory of Convex Structures that may be of interest. First define a closure system. Suppose that $X$ is a set and $\mathcal{C}$ is a collection of subsets of $X$. Then $\langle X, \mathcal{C} \rangle$ is a closure system if and only if for all $\mathcal{A} \subseteq \mathcal{C}$ we have $\cap \mathcal{A} \in \mathcal{C}$. Sometimes people require $\varnothing \in \mathcal{C}$. I can't remember if van de Vel does. For $A \subseteq X$ define $$\mathsf{cl}(A) = \cap \{ C \in \mathcal{C} \colon A \subseteq C \} .$$ In any event the convex subsets of a real vector space satisfy these properties.
Theorem Suppose that $\langle X, \mathcal{C} \rangle$ is a closure system. Then the following statements are equivalent:
- For all $A \subseteq X$ we have $\mathsf{cl}(A) = \cup \{ \mathsf{cl}(F) \colon F \text{ is a finite subset of } A \} $.
- For all $\mathcal{D}$ which are collections of subsets of $X$ that are directed by inclusion (see below for a definition of directed by inclusion) we have $\mathsf{cl}(\cup \mathcal{D}) = \cup \{ \mathsf{cl}(D) \colon D \in \mathcal{D} \} $.
- For all $\mathcal{T}$ which are collections of subsets of $X$ that are totally ordered by inclusion (see below for a definition of totally ordered by inclusion) we have $\mathsf{cl}(\cup \mathcal{T}) = \cup \{ \mathsf{cl}(T) \colon T \in \mathcal{T} \} $.
A collection $\mathcal{D}$ of subsets is directed by inclusion if and only if for all $D_{0}, D_{1} \in \mathcal{D}$ there is a $D \in \mathcal{D}$ with $D_{0}, D_{1} \subseteq D$.
A collection of subsets is totally ordered by inclusion if and only if for all $T_{0}, T_{1} \in \mathcal{T}$ we have $T_{0} \subseteq T_{1}$ or $T_{1} \subseteq T_{0}$.
Edit
This is in response to Tim's comment.
Suppose that $X$ is a set and $\mathcal{F}$ is the collection of finite subsets of $X$. Suppose also that $f \colon \mathcal{F} \rightarrow \mathcal{P}(X)$ where $\mathcal{P}(X)$ is the collection of all subsets of $X$. For each $n \in \mathbb{N}$ define
\begin{align}
g_{n} \colon \mathcal{P}(X) &\rightarrow \mathcal{P}(X) \\
\text{for all $A \subseteq X$ by the assignment} \\
g_{0} \colon A &\mapsto A \\
g_{n} \colon A & \mapsto g_{n-1}(A) \cup (\cup \{ f(F) \colon F \subseteq g_{n-1}(A) \text{ is finite} \} )
\end{align}
Then $A \mapsto \cup \{ g_{n}(A) \colon n \in \mathbb{N} \} $ is a convex hull operator. The function $f$ generalizes the notion of the points between $x, y \in \mathbb{R}^{k}$. If you think of this construction in this manner then the $g_{n}$ functions just accumulate line segments.
Best Answer
For simplicity, let us denote $T$ as the operation of taking the convex hull of a set.
We observe following basic principles.
$A \subset T(A)$.
If $A \supset B$, then $T(A) \supset T(B)$.
$T(T(A))=T(A)$.
Now we prove the proposition.
From 2, $T(A \cup B) \supset T(A)$ and also $T(A \cup B ) \supset T(B)$, so $T(A \cup B) \supset T(A) \cup T(B)$. Hence $T(A \cup B) \supset T(T(A) \cup T(B))$.
On the other hand, we have $ A \cup B \subset T(A) \cup T(B)$. Therefore, $T(A \cup B) \subset T(T(A) \cup T(B))$.