Let $a_1,a_2,\ldots,a_k$ and $p$ are positive real numbers where $a_1$ is minimum. If we solve $$\frac{1}{x-a_1}+\frac{1}{x-a_2}+\ldots \frac{1}{x-a_k}-1=0 \quad(1)$$ we have a $k$-degree polynomial.
Consider another polynomial, given by: $$\frac{1}{x-a_1}+\frac{1}{x-a_2}+\ldots \frac{1}{x-a_k}-\frac{p}{x}-1=0\quad(2)$$ I have two observation:
$(a)$ Least root of $(1)$ lies in $(a_1,a_1+1).$
$(b)$ Least root of $(2)$ is less than that of $(1).$
Are the two observations true? If so, how to prove them?
Let $A=\;\{a_j \mid j = 1, 2, \cdots k\}\;$ and $\;f(x)=\sum_{j=1}^k \frac{1}{x-a_j} - 1$. Observe that:
$f'(x) = \sum_{j=1}^k \frac{-1}{(x-a_j)^2} \;\;\lt\;\; 0\;$ so $f$ is strictly decreasing on $\mathbb{R} \setminus A$
$f$ has vertical asymptotes at each $a_j \in A$ with lateral limits $f(a_j^-)=-\infty$, $f(a_j^+)=+\infty$
$f(x) \lt 0$ for $x \lt a_1$ so there are no roots in $(-\infty, a_1)$
Therefore on each interval $(a_j,a_{j+1})$ the function monotonically decreases from plus to minus infinity, so there exists exactly one root $x_j \in (a_j,a_{j+1})$ for $1 \le j \lt k$, and the smallest root is $x_1$.
If $a_2 \le a_1 + 1$ then $x_1 \in (a_1,a_2) \subseteq (a_1, a_1+1)$ satisfies (a).
If $a_2 \gt a_1+1$ then $f(a_1+1)=\sum_{j=2}^k \frac{1}{a_1-a_j+1} \;\lt\;0$ because $a_1 - a_j+1 \le a_1 - a_2+1 \lt 0$. Since $f(a_1^+)=\infty$ and $f(a_1+1) \lt 0$ it follows that $x_1 \in (a_1,a_1+1)$ which concludes the proof.
Let $g(x)= f(x) - \frac{p}{x}$ and assume $p \ne 0$, otherwise $g \equiv f$ and (b) does not hold.
If $p \gt 0$ observe that $\lim_{x \to -\infty} g(x) = -1$ and $\lim_{x \to 0^-} g(x) = +\infty$. It follows that $g(x)$ has a root $x_0 \lt 0 \lt x_1$.
If $p \lt 0$ observe that $\lim_{x \to 0^+} g(x) = +\infty$ and $\lim_{x \to a_1^-} g(x) = -\infty$. It follows that $g(x)$ has a root $x_0 \lt a_1 \lt x_1$.