How to formulate the Lagrange dual problem for a convex optimisation? Formulate the Lagrangian dual problem for the explicit convex problem below:
$$C=\{(x,y) \mid x^4 + y^4 + 2(x^2y^2-x^2-y^2) -8 \leq 0\}$$
The original problem was:
$$min: x - y$$
$$s.t. x,y \in C$$
Then converted into:
$$min: x - y$$
$$s.t.:  x^4 + y^4 + 2(x^2y^2-x^2-y^2) -8 \leq 0$$
Which I've then rewritten as:
$$inf_{(x,y \in R)} sup_{z\in R}(x-y + z[x^4 + y^4 + 2(x^2y^2-x^2-y^2) -8] )$$
which would mean the dual function is:
$$g(z)=inf_{(x,y \in R)}(x-y + z[x^4 + y^4 + 2(x^2y^2-x^2-y^2) -8])$$
But this is where ive gotten stuck again.
I haven't seen any examples of this type, feel free to point me in the right direction if you have already answered elsewhere.
 A: Construct the Lagrangian: $$L(x,y;\lambda)=x-y+\lambda(x^4+y^4+2(x^2y^2-x^2-y^2)-8) $$
Then we look for the minimum of $L$ with respect to $x,y$. For that we will need the derivative to be equal to zero: $$\frac{\partial L}{\partial x}= 1+\lambda(4x^3+4xy^2-4x)=0 \iff \\ 4\lambda x^3+4\lambda(y^2-1)x+1=0$$
where similarly, the equation with respect to $y$ is $4\lambda y^3+4\lambda(x^2-1)y-1=0$.
These both look like nasty equations, but fortunately we a formula to solve a cubic equation. There might be a problem if you have multiple real roots (there can be up to 3), because we might not be able to know which of the roots actually minimizes the $L$ function. However, I assume that the problem was constructed such that the cubic equations  only have one root. I will leave that part for you.
Now that you have found the optimal $x^*, y^*$ plug them back into $L(x^*,y^*;\lambda)=\inf_{x,y} L(x,y;\lambda)=q(\lambda)$ and you have your dual function. Then the dual problem is $$\max_{\lambda\geq 0} q(\lambda)$$
