Show $\delta_{\prod(\mu,\nu)}(\pi) = \sup_{\varphi,\psi} [\int \varphi d\mu + \int \psi d\nu - \int \varphi d\pi - \int \psi d\pi]$

We let $X \sim \mu$ and $Y \sim \nu$ be the source space and target space for optimal transport. And we let $c: X \times Y \rightarrow \mathbb{R} \cup \{\infty\}$ be a cost function. We recall that $\pi \in \prod(\mu,\nu)$ is a coupling of $\mu,\nu$, if $\text{proj})_{\#}^{X}(\pi) = \mu, \text{proj}_{\#}^{Y}(\pi) = \nu$.
We define $$\delta_{\prod(\mu,\nu)}(\pi) = \begin{cases} 0, & \text{ if } \pi \in \prod(\mu,\nu)\\ \infty, &\text{ otherwise }\end{cases}.$$ We want to show $$\sup_{\varphi,\psi} \int \varphi d\mu + \int \psi d\nu - \int \varphi d\pi - \int \psi d\pi = \delta_{\pi (\mu,\nu)}(\pi),$$ where $\varphi$ and $\psi$ satisfy the constraint $\varphi(x) + \psi(y) \leq c(x,y)$ for all $(x,y) \in X \times Y$.
My proof so far: We first consider the case when $\pi \in \prod(\mu,\nu)$. When $\pi \in \prod(\mu,\nu)$, we have $\delta_{\prod(\mu,\nu)}(\pi) = 0$ by construction. This means that we need to show $\sup_{\varphi,\psi} \int \varphi d\mu + \int \psi d\nu - \int \varphi d\pi - \int \psi d\pi = 0$.
We define $\varphi \oplus \psi$ be a function on $X \times Y$ by $(\varphi \oplus \psi)(x,y) = \varphi(x) + \psi(y)$. Then if $\pi \in \prod(\mu,\nu)$, we have $$\int (\varphi \oplus \psi) d\pi = \int \varphi(x) d\pi (x,y) + \int \psi (y) d\pi (x,y) = \int \varphi d\mu + \int \psi d\nu.$$ So we have when $\pi \in \prod(\mu,\nu)$, $$\int \varphi d\mu + \int \psi d\nu - \int (\varphi + \psi)d\pi = 0.$$ So when $\pi \in \prod(\mu,\nu)$, we have $\sup_{\varphi,\psi} \int \varphi d\mu + \int \psi d\nu - \int \varphi d \pi - \int \psi d\pi = 0$.
However, I’m not sure how to deal with the case when $\pi \notin \prod(\mu,\nu)$.