Consider the following splitting problem. Given $Y$ balls of which $X\leqslant Y$ of them are blue balls. The goal is to split the balls by placing them in $K$ baskets based on the following quadratic-over-linear program: $$ \begin{array}{cr} \displaystyle\min_{\mathbf{x},\mathbf{y}\in\mathbf{Z}_{+}^{K}}&\displaystyle\mathbf{x}^{T}\big(\mathbf{P}(\mathbf{y})\big)^{-1}\mathbf{x} \\ \mathrm{s.t.}&\mathrm{tr}(\mathbf{P}(\mathbf{y}))=Y \\ &\|\mathbf{x}\|_{1}=X \\ & \mathbf{x}\preceq\mathbf{y} \end{array} $$ In my case, the matrix $\mathbf{P}(\mathbf{y})$ is a diagonal matrix. For example, for $K=3$ baskets we have: $$ \mathbf{P}(\mathbf{y}):=\begin{bmatrix}y_{1} & 0 & 0 \\ 0 & y_{2} & 0 \\ 0 & 0 & y_{3} \end{bmatrix} $$
Observation:
If $\gcd(X,Y)=1$, my conjecture is that the global solution for this splitting problem for $K$ baskets can be recursively found using the global solution for a sub-problem with $K-1$ baskets. To illustrate better, consider $X=41$ and $Y = 100$. Thus, the global solution for increasing $K$ is given by: $$ \begin{array}{ccll} &K&\mathbf{x}&\mathbf{y} \\ &2&\begin{bmatrix}16&\color{red}{25} \end{bmatrix}&\begin{bmatrix}39&\color{red}{61}\end{bmatrix} \\ &3& \begin{bmatrix}16& {\color{red}9}&\color{red}{16} \end{bmatrix} &\begin{bmatrix}39&\color{red}{22}&\color{red}{39}\end{bmatrix} \\ (\text{re-arranged}) & 3& \begin{bmatrix}9&16&\color{red}{16} \end{bmatrix} &\begin{bmatrix}22&39&\color{red}{39}\end{bmatrix} \\ &4& \begin{bmatrix}9& 16& \color{red}{7} & \color{red}{9} \end{bmatrix} &\begin{bmatrix}22&39&\color{red}{17} & \color{red}{22}\end{bmatrix} \\ (\text{re-arranged}) &4& \begin{bmatrix}7 & 9 & 9& \color{red}{16} \end{bmatrix} & \begin{bmatrix}17 & 22 & 22&\color{red}{39}\end{bmatrix} \\ & \vdots \end{array} $$ It appears that if I increase the number of baskets, then the global solution for the next problem is taking the basket with the largest number of balls (as highlighted by red), i.e. $y^{(j)}:=\max_{j=1,2,...}\mathbf{y}$ and splitting it into two baskets as a global solution for the sub-problem: $$ \begin{array}{cl} \displaystyle\min_{\widehat{\mathbf{x}},\widehat{\mathbf{y}}\in\mathbf{Z}_{+}^{2}}&\displaystyle\begin{bmatrix}x_{1}^{(j)} & x_{2}^{(j)} \end{bmatrix} \begin{bmatrix}y_{1}^{(j)} & 0 \\ 0 & y_{2}^{(j)} \end{bmatrix}^{-1} \begin{bmatrix}x_{1}^{(j)} \\ x_{2}^{(j)} \end{bmatrix} \\ \mathrm{s.t.}& x_{1}^{(j)} + x_{2}^{(j)} = x^{(j)}\\ & y_{1}^{(j)} + y_{2}^{(j)} = y^{(j)} \\ &\widehat{\mathbf{x}}\preceq\widehat{\mathbf{y}} \end{array} $$ Then, we concatenate this to the global solution for the problem with $K-1$ baskets.
For example, take $X=41$, $Y=100$, and $K=4$.
The global solution for the previous problem, i.e. for $X=41$, $Y=100$, $K=3$, is $\mathbf{x}=\begin{bmatrix}9 & 16 & 16 \end{bmatrix}$ and $\mathbf{y}=\begin{bmatrix}22 & 39 & 39 \end{bmatrix}$.
Take the basket with largest amount of blue balls $\mathbf{x}$ (or balls in general $\mathbf{y}$), which in this case is $x=16$ and $y=39$, and optimally split it to two based on the sub-problem above. We get $\widehat{\mathbf{x}}=\begin{bmatrix}7&9\end{bmatrix}$ and $\widehat{\mathbf{y}}=\begin{bmatrix}17&22\end{bmatrix}$
Obtain $\mathbf{x}^{\star}=\begin{bmatrix}9&16&\widehat{\mathbf{x}}\end{bmatrix} = \begin{bmatrix}9&16& 7 & 9\end{bmatrix}$ and $\mathbf{x}^{\star}=\begin{bmatrix}9&16&\widehat{\mathbf{y}}\end{bmatrix} = \begin{bmatrix}22&39& 17 & 22\end{bmatrix}$$\mathbf{y}^{\star}=\begin{bmatrix}9&16&\widehat{\mathbf{y}}\end{bmatrix} = \begin{bmatrix}22&39& 17 & 22\end{bmatrix}$
Questions:
Why does this property hold for $\gcd(X,Y)=1$? I don't know where to start to prove the separability of this problem. I couldn't find much in literature about this (or a generalization) type of problem.
What is a more abstract way to define an operator that takes you from the problem of $K$ baskets to the problem with $K+1$ basket?