Let $\lambda_1 \ge \lambda_2 \ge \ldots \ge \lambda_d$ be positive numbers and let $0 \lt c \le d$.
Question. What is a good upper-bound for the positive solutions of the equation $$ \sum_{j=1}^d \frac{1}{x+\lambda_j} = c. \tag{1} $$
Note that if we rewrite the above equation as $$ \tag{2} c x^d + a_{d-1}x^{d-1} + \ldots a_1 x + a_0=0, $$ for some $a_{d-1},\ldots,a_0 \in \mathbb R$ with $a_d=c$ and $a_0=c\Pi_j\lambda_j - \sum_j \lambda_j$. then an upper-bound is given by $$ B := 2\cdot \max_{a_j a_d \lt 0} |a_j/a_d|^{1/2}. \tag{3} $$
See this wikipedia page https://en.wikipedia.org/wiki/Geometrical_properties_of_polynomial_roots#Bounds_of_positive_real_roots.
However, I don't see a simple way to write the $a_j$'s a function of the $\lambda_j$'s, for $1 \le j \le d-1$, in order to exploit (3).
Denote $x_0$ be the largest positive root and $p_i = \frac{1}{\lambda_i}$ for $i=1,...,d$, then
$$c = \sum_{i=1}^d \frac{1}{x_0+\lambda_i} =\sum_{i=1}^d \frac{p_i}{p_ix_0+1} \tag{1}$$
The function $f(p) = \frac{p}{px_0+1}$ is concave over the domain $p>0$ because $$f''(p) = -\frac{2x_0}{(1+px_0)^3} < 0$$ Then we can apply the Jensen's inequality to $(1)$ $$c=\sum_{i=1}^d f(p_i) \le d\cdot f\left(\frac{\sum_{i=1}^d p_i}{d}\right ) = d\cdot \frac{\sum_{i=1}^d p_i}{x_0\sum_{i=1}^d p_i+d} = \frac{d}{x_0+\frac{d}{\sum_{i=1}^d p_i}}$$ $$\Longleftrightarrow x_0 \le \color{red}{\frac{d}{c} -\frac{d}{\sum_{i=1}^d \frac{1}{\lambda_i}} }$$