There is a finite real set $S=\{\vec{v_1}, \cdots, \vec{v_i},\cdots, \vec{v_n}\}$ and $\vec{v_i} \in R^{m}$ and a convex function $f(\vec{a}, \vec{b}): R^m \times R^m \to R$.
Define a 2 way partition as $A \cup B = S$ and $A \cap B = \emptyset $.
$\vec{a} = \sum_{i \in A} \vec{v_i}$ and $\vec{b} = \sum_{i \in B} \vec{v_i}$.
An optimal partition $A^*, B^*$ that maximize $f(\vec{a}, \vec{b})$.
And $F(S) := (A^*, B^*)$
Define a 3 way partition as $\cup(X, Y, Z) = S$ and $ \cap(X, Y, Z) = \emptyset $.
$\vec{x} = \sum_{i \in X} \vec{v_i}, \vec{y} = \sum_{i \in Y} \vec{v_i}, \vec{z} = \sum_{i \in Z} \vec{v_i}$.
Define $g(\vec{x}, \vec{y}, \vec{z}) = |X \cup Y\cup Z| *f(\vec{x}, \vec{y}+\vec{z}) +|Y\cup Z|* f(\vec{y}, \vec{z})$.
An optimal partition $X^*, Y^*, Z^*$ that maximize $g(\vec{x}, \vec{y}, \vec{z})$.
Is $F(Y^* \cup Z^*)$ the same as $(Y^*, Z^*)$ ?
And would those hold $F(X^* \cup Y^*) == (X^*, Y^*), F(X^* \cup Z^*)==(X^*, Z^*)$ ?
$F(\cup(Y^*, Z^*))$ is the same as $(Y^*, Z^*)$. It can be proved easily with proof by contradiction.
If $F(\cup(Y^*, Z^*)) != (Y^*, Z^*)$, there exists an optimal partition of $F(\cup(Y^*, Z^*)) = (\hat{Y}, \hat{Z})$, such that $(\hat{Y}, \hat{Z}) != (Y^*, Z^*)$ and $f(\vec{y^*} ,\vec{z^*}) < f(\vec{\hat{y}} , \vec{\hat{z}})$
And $\cup(Y^*, Z^*) = \cup(\hat{Y}, \hat{Z})$, which leads to $\vec{y^*} + \vec{z^*} = \vec{\hat{y}} + \vec{\hat{z}}$
$$g(\vec{x^*}, \vec{y^*}, \vec{z^*}) = |S|* f(\vec{x^*}, \vec{y^*}+\vec{z^*})+|\cup(Y^*, Z^*)|*f(\vec{y^*} ,\vec{z^*}) \\ = |S|* f(\vec{x^*}, \vec{\hat{y}} + \vec{\hat{z}}) + |\cup(\hat{Y}, \hat{Z})|*f(\vec{y^*} ,\vec{z^*}) \\ < |S|* f(\vec{x^*}, \vec{\hat{y}} + \vec{\hat{z}}) + |\cup(\hat{Y}, \hat{Z})|*f(\vec{\hat{y}} , \vec{\hat{z}})\\ = g(\vec{x^*}, \vec{\hat{y}} , \vec{\hat{z}})$$.
Which contradicts with $(\vec{x^*}, \vec{y^*}, \vec{z^*})$ maximize $g$. So that $F(\cup(Y^*, Z^*)) == (Y^*, Z^*)$