Dual optimization problem for standard linear programming Let $K$ be a proper convex cone. Let's have a cone problem
$$
\text{min } \textbf{c}^T\textbf{x} \\ \text{s.t. } A\textbf{x} =\textbf{b}\\\textbf{x}\succeq_K \textbf{0}
$$
How to derive the dual program for this one? Thx
I know that when solving duals one works with the Lagrange dual function - but in this case, I have troubles because of the cone inequality.
 A: Notice: 

If $K$ is a proper cone, we have:
$$\mathbf{x} \succeq_{K} \mathbf{y} \text{ if and only if }
\boldsymbol{\lambda}^T \mathbf{x} \geq \boldsymbol{\lambda}^T
\mathbf{y} \text{ for all } \boldsymbol{\lambda} \in K^*,$$ where
  $K^*$ is the dual cone of $K$.

Suppose the generalized inequality is the component wise inequality defined by $K = \mathbb{R}_+^n$. For the standard linear programming (LP) problem we have: 
\begin{aligned}
    & \min_{\mathbf{x} \in \mathbb{R}^n} && \hspace{-1mm} \mathbf{c}^T \mathbf{x} \\
    & \hspace{2mm} \text{s.t.}  && \hspace{-4mm} \mathbf{A}\mathbf{x} = \mathbf{b} \\
    & \text{}       && \hspace{-2mm} \mathbf{x}  \succeq \mathbf{0}
\end{aligned}
The Lagrangian can be written as
\begin{aligned}
\mathcal{L}(\mathbf{x},\boldsymbol{\lambda},\boldsymbol{\nu}) &= \mathbf{c}^T \mathbf{x} - \boldsymbol{\lambda}^T \mathbf{x}+ \boldsymbol{\nu}^T(\mathbf{A}\mathbf{x}-\mathbf{b}) \\
&= (\mathbf{c}-\boldsymbol{\lambda}+\mathbf{A}^T\boldsymbol{\nu})^T \mathbf{x} - \mathbf{b}^T \boldsymbol{\nu}.
\end{aligned}
As it is clear $\mathcal{L}$ is an affine function over $\mathbf{x}$ and hence it is unbounded below except for the case $\mathbf{c} \boldsymbol{\lambda}+\mathbf{A}^T\boldsymbol{\nu} = \mathbf{0}$. Thus, the dual function is 
\begin{aligned}
g(\boldsymbol{\lambda},\boldsymbol{\nu}) &= \inf_{\mathbf{x}} \mathcal{L}(\mathbf{x},\boldsymbol{\lambda},\boldsymbol{\nu}) \\
&= \begin{cases}
     - \mathbf{b}^T \boldsymbol{\nu}  & \quad \text{if:} \;\mathbf{c}-\boldsymbol{\lambda}+\mathbf{A}^T\boldsymbol{\nu} = \mathbf{0}\\
    -\infty  & \quad \text{otherwise}
  \end{cases},
\end{aligned}
where $\text{dom } g = \{(\boldsymbol{\lambda},\boldsymbol{\nu})| \; \mathbf{c}-\boldsymbol{\lambda}+\mathbf{A}^T\boldsymbol{\nu} = \mathbf{0}\}$.
Then, the Lagrange dual problem of the LP problem can be obtained as 
\begin{aligned}
    & \hspace{-10mm} \max_{(\boldsymbol{\lambda},\boldsymbol{\nu}) \in \text{dom } g} && \hspace{-1mm} g(\boldsymbol{\lambda},\boldsymbol{\nu}) \\
    & \hspace{-3mm} \text{s.t.}  && \hspace{-1mm} \boldsymbol{\lambda}  \succeq \mathbf{0},
\end{aligned}
and hence the dual problem can be described as
\begin{aligned}
   & \hspace{6mm}  \max_{\boldsymbol{\lambda},\boldsymbol{\nu}} && \hspace{-5mm}  - \mathbf{b}^T \boldsymbol{\nu}  \\
    & \hspace{8mm} \text{s.t.}  && \hspace{-3mm} \boldsymbol{\lambda}  \succeq \mathbf{0} \\ 
& {} && \hspace{-3mm} \mathbf{c}-\boldsymbol{\lambda}+\mathbf{A}^T\boldsymbol{\nu} = \mathbf{0}.
\end{aligned}
