So if I understand well, your decision variable is finally a vector. Your problem is thus

\begin{array}{cl}
\min_{\mu\in\mathbb{R}^n} & \sum_{i=1}^n \frac{1}{\mu_i-\lambda_i} \\
s.t. & \sum_{i=1}^n \frac{1}{\mu_i} \leq c \\
& \lambda_i < \mu_i
\end{array}

If by any chance all your \lambda_i are positive, I can propose the following. Perform the change of variable t_i = 1/\mu_i, so that your problem is equivalent to

\begin{array}{cl}
\min_{t\in\mathbb{R}^n} & \sum_{i=1}^n \frac{t_i}{1-\lambda_i t_i} \\
s.t. & \sum_{i=1}^n t_i \leq c \\
& 0 < t_i < 1/\lambda_i
\end{array}

Introduce now the vector x, the problem is still equivalent to

\begin{array}{cl}
\min_{t\in\mathbb{R}^n,x\in\mathbb{R}^n} & \sum_{i=1}^n x_i \\
s.t. & \sum_{i=1}^n t_i \leq c \\
& 0 < t_i < 1/\lambda_i \\
& 0 < x_i \\
& \frac{t_i}{1-\lambda_i t_i} \leq x_i
\end{array}

that is (1-\lambda_i t_i and x_i being positive)

\begin{array}{cl}
\min_{t\in\mathbb{R}^n,x\in\mathbb{R}^n} & \sum_{i=1}^n x_i \\
s.t. & \sum_{i=1}^n t_i \leq c \\
& 0 < t_i < 1/\lambda_i \\
& 0 < x_i \\
& \frac{t_i}{x_i} \leq 1-\lambda_i t_i
\end{array}

Since the t_i are positive, this is equivalent to

\begin{array}{cl}
\min_{t\in\mathbb{R}^n,x\in\mathbb{R}^n} & \sum_{i=1}^n x_i \\
s.t. & \sum_{i=1}^n t_i \leq c \\
& 0 < t_i < 1/\lambda_i \\
& 0 < x_i \\
& \frac{t_i^2}{x_i} \leq t_i-\lambda_i t_i^2
\end{array}

which is convex.

From a coding point of view, I think that quadoverlin understands that the x_i should be positive. And you might prefer the constraint 1/t_i < \lambda_i using invpos.