If $X := \left\{x \in \mathbb{R}^n \mid Px \leq p \right\}$ and $Y := \left\{ y \in \mathbb{R}^m \mid Qy \leq q \right\}$ are nonempty and bounded polyhedral sets, and if $a\in \mathbb{R}^n, b\in\mathbb{R}^m$, and $A\in\mathbb{R}^{m\times n}$, how do we prove the following?
$$\min_{x \in X} \max_{y \in Y} \left\{ a^T x + b^T y + y^T Ax \right\} = \max_{y \in Y} \min_{x \in X} \left\{a^T x + b^T y + y^T Ax \right\}$$
I can prove that $\min_{x\in X} \max_{y\in Y} \{a^T x + b^T y + y^T Ax \} \geq \max_{y\in Y} \min_{x\in X} \{a^T x + b^T y + y^T Ax \}$ by considering a specific $(x', y') \in X\times Y$. Specifically, letting $f(x, y) = a^T x + b^T y + y^T Ax $, we have \begin{align*} f(x', y') &\geq \min_{x\in X} f(x, y')\\ \implies \max_{y\in Y} f(x', y) &\geq \max_{y\in Y} \min_{x\in X} f(x, y)\\ \implies \min_{x\in X}\max_{y\in Y} f(x, y) &\geq \max_{y\in Y} \min_{x\in X} f(x, y). \end{align*}
However, I am not sure how to continue to show the reverse inequality, namely to show that $\min_{x\in X} \max_{y\in Y} \{a^T x + b^T y + y^T Ax \} \leq \max_{y\in Y} \min_{x\in X} \{a^T x + b^T y + y^T Ax \}$.
My intuition says we need to construct a dual LP and use weak duality, but that hasn't worked. Could someone nudge me in the right direction? I would greatly appreciate it.
