This is a question about what forms of new Lagrangian are valid after we alter an original Lagrangian to add Lagrange multipliers.
Say we have some constraint $g(x,y)=0$ that we want to add to our Lagrangian $L(x,y)$ using Lagrange multipliers to form a new Lagrangian $\bar{L}(x,y,\lambda)$ as
$$ \bar{L}=L(x,y)+\lambda g(x,y) \label{1}\tag{1}$$
My understanding here is that, whenever we add such a constraint, it has to be such that the new and old Lagrangian are the same if we directly impose out knowledge that $g=0$. That is, equation \eqref{1} above has $g=0$ and therefore we have not fundamentally altered $L$. If we then carry on with the optimization with $\lambda$ included as variable, we get the usual treatment of Lagrange multipliers.
Now let's say the constraint is of the form $x=\bar{x}$ and $y=\bar{y}$, where $\bar{x},\bar{y}$ is another pair of variables, we can then talk about a new Lagrangian of the form $$ \bar{L}=L(x,y)+\lambda_{1}(x-\bar{x})+\lambda_{2}(y-\bar{y}), \label{2}\tag{2}$$ which, again, seems OK to me, because we wouldn't fundamentally alter the old $L$ by this formulation.
My question now is, can we add another term that will alter $L$ to be written mixed variables like: $$ \bar{L}=L(x,y)+L(\bar{x},\bar{y})+\lambda_{1}(x-\bar{x})+\lambda_{2}(y-\bar{y}), \label{3}\tag{3}$$ where the first two terms on the RHS are the same function but in different variables.
My initial feeling is that this should work, because if we enforce $x=\bar{x}$ and $y=\bar{y}$ then the the Lagrangian will be the same old form, except for a scalar multiplier of \eqref{2}, which shouldn't alter the dynamics. Does that make sense?