Suppose we have the following two convex optimization problems:
$$\max_{x_i} \; \sum_{i \in I} w_i \cdot f_i(x_i) \\ \text{subject to:} \quad 0 \leq x_i \leq 1, \quad \forall i \in I \\ \sum_{i \in I} x_i \leq c$$
and given that we reserve $x_i$ beforehand:
$$\max_{x_{i '}} \; \sum_{i ' \neq i} w_{i '} \cdot f_{i '}(x_{i '}) \\ \text{subject to:} \quad 0 \leq x_{i '} \leq 1, \quad \forall i ' \neq i \\ \sum_{i ' \neq i} x_{i '} \leq c - x_i,$$
where $f_i$ is concave, strictly increasing, and differentiable; $c$ is a positive real number, $w_i$ is a given positive real number for all $i \in I$, and $I = \{1, 2, \dots, n\}$.
By Slater's condition (see Boyd's "Convex Optimization"), both optimization problems have a unique and optimal solution, denoted by $x ^ *$ and $x_{- i} ^ *$, if and only if they satisfy all the constraints of their respective problem and there exist a set of Lagrange multipliers, say $\lambda$ for the first problem and $\mu$ for the second, that satisfy the KKT conditions.
My question is the following: Given $x_{- i} ^ *$ and $\mu$, can we find a bound on $w_i$ such that $(x_i, x_{- i})$ satisfies the KKT of the first problem. For example, I thought that maybe if we set $$w_i \geq \frac{\mu}{\frac{d f_i(x_i)}{d x_i}}$$ then we can get a necessary and sufficient condition $$\mu \geq \lambda$$ for $(x_i, x_{- i})$ to be the unique solution of the first problem.
In other words, can one show a relation between two Lagrange multipliers of similar convex optimization problems?